An Odd Ratio

The sum $S$ of an arithmetic progression of $n$ terms is given by: $S=\dfrac{n}{2} (a+l)$ , where $a$ and $l$ are the first and last terms respectively.

Therefore,

$\dfrac {100001+100003+100005+...+199999} {1+3+5+7+9+11...+99999}$

$= \dfrac {\frac{n}{2}(a_1+l_1)} {\frac{n}{2}(a_2+l_2)} = \dfrac {\frac{50000}{2}(100001+199999)} {\frac{50000}{2}(1+99999)} = \dfrac {300000}{100000} = \boxed{3}$

Notice that we can form $25000$ pairs of numbers whose sum is $100000$ in the denominator: $(1 + 99999), (3 + 99997), (5 + 99995)$ etc.. And in the numerator we can do the same thing, but $100000$ is added to each of the $50000$ terms. So we have:

$\frac {100001 + 100003 + 100005 + \ldots + 199999}{1 + 3 + 5 + 7 + 9 + 11 + \ldots + 99999}$

$= \frac {25000 \times 100000 + 50000 \times 100000}{25000 \times 100000}$

$= \frac {75000 \times 100000}{25000 \times 100000}$

$= \frac {75000}{25000} = \frac {75}{25} = \boxed{3}$

If we define $n$ to be the number of terms within the divisor. Then if we look at $n^{th}$ term, we have: $\frac { (2n+1)+(2n+3)+...+(4n-3)+(4n-1) }{ 1+3+5+...+(2n-3)+(2n-1) }$ By inputting the values $n=1, n=2$ and $n=3$ , we achieve the following sequences: $\begin{matrix} n= \\ Divisor: \end{matrix}\begin{matrix} 1 & 2 & 3 \\ 1 & 4 & 9 \end{matrix}$ Clearly we can see that for the divisor, the $n^{th}$ term equals $n^2$

Repeating the process for the dividend, we get: $\begin{matrix} n= \\ Dividend: \end{matrix}\begin{matrix} 1 & 2 & 3 \\ 3 & 12 & 27 \end{matrix}$ Then, by finding the $n^{th}$ term of the sequence using a method of your liking, you should end up with an $n^{th}$ term of $3n^2$

Now, by placing our two $n^{th}$ term values back into the original equation, we have: $\frac { 3n^2 }{ n^2 }$ By simplifying this equation we end up with an answer of $\boxed { 3 }$

Note: The equation shown is for $n = 50000$ , which would still result to 3.

the given expression can be written as

$\frac { (100000+1)+(100000+3)+(100000+5)+.....(100000+99999) }{ 1+3+5.....+99999 }$

now separating $100000$ we get the expression as

$\frac { (100000\times 50000)+(1+3+5....+99999) }{ 1+3+5...+99999 } \quad (\because \quad no.\quad of\quad terms\quad are\quad 50000)$

now separating the fraction as $\frac { (100000\times 50000) }{ 1+3+5...+99999 } +\frac { (1+3+5....+99999) }{ 1+3+5...+99999 }$

we get $\frac { 100000\times 50000 }{ \frac { 50000 }{ 2 } (1+99999) } +1$

finally the answer is $1+2$ = $\boxed{3}$

I hope this explanation will be easy to understand. First off , we can see that 100,001+199,999=300,000. Similarly 100,003+199,997=300,000 and so on.This is the same with the denimenator.Since the number of terms being added in the numerator and the denominator are the same and give the same 300,000/100,000=3,3 is the answer.

There's actually a simple solution to it. The denominator is the sum of the first x odd numbers. The numerator is the sum of the next x odd numbers, or the sum of the first 2x, minus the sum of the first x. The result is $x^2$ for the denominator and $(2x)^2 - x^2$ , $4x^2 - x^2$ , or $3x^2$ for the numerator.

Thus, as long as the number of terms is the same, and the numerator picks up where the denominator left off, the answer is always 3.

100001 + 100003 + 100005 + ... + 199999

= (100000+1) + (100000+3) + (100000+5) + ... + (100000+99999)

= (100000 + 100000 + 100000 + ... + 100000) + (1 + 3 + 5 + 7 + ... + 99999)

= (1+99999) + (2+99998) + (3+100000) + ... + (99999+1) + (1 + 3 + 5 + 7 + ... + 99999)

= 3 × (1 + 3 + 5 + 7 + ... + 99999)

$1 + 3 + 5 + 7 + ... + n = (\frac{n+1}{2}) ^2$ , so the denominator sums to $(\frac{99999+1}{2}) ^2 = 50000^2$ .

Whereas, for the numerator, we have $(\frac{199999+1}{2}) ^2 - 50000^2 = 100000^2 - 50000^2 = 3 \times50000^2$ .

$\large\frac{100001+100003+100005+\ldots+199999}{1+3+5+7+9+11+\ldots+99999}$

$\large=\frac{100000+1+100000+3+100000+5+\ldots+100000+99999}{1+3+5+7+9+11+\ldots+99999}$

$\large=\frac{(100000+100000+100000+\ldots+100000)+(1+3+5+\ldots+99999)}{1+3+5+7+9+11+\ldots+99999}$

$\large=\frac{(n_i\times100000)+(1+3+5+\ldots+99999)}{1+3+5+7+9+11+\ldots+99999}$

$\large=\frac{n_i\times100000}{1+3+5+\ldots+99999}+1$

no of integers $\space\large n_i= \frac{99999-1}{2}+1=50000$

Sum of all odd integers of n (odd) numbers $\space\large\color{#3D99F6}{=\frac{n-1}{4}*(n+1)+\frac {n+1}{2}}$

$\large\color{#3D99F6}{=\frac{n^2-1+2(n+1)}{4}}$

$\large\color{#3D99F6}{=\frac{99999^2-1+2(99999+1)}{4}=2500000000=2.5\times 10^9}$

$\large=\frac{5\times10^9}{2.5\times 10^9}+1=2+1=\boxed{\color{#D61F06}{3}}$

well their common difference is 2 (both the numerator and denominator) and the difference from the first and last term (for both the numerator and the denominator) is the same: 99998. therefore, the number of terms for both numerator and denominator is the same; and so, just get the sum of the first and last terms of the numerator and divide it by the first and last terms of the denominator. There is no need for solving the exact values of the numerator and denominator since the number of terms are the same. You will just end up cancelling them out if you did.

Equation: (100001+199999) / （1+99999) = 3

Take 100001=100000+1 and so on till 199999.We get 100000*50000+(50000)^2/50000^.This solving we get 3

Let $a = 100000$ . The expression now becomes:

$\frac{(a+1)+(a+3)+(a+5)+\dotsb+(a+99999)}{1+3+5+7+\dotsb+99999}.$

We should note now, that between $1$ and $99999$ there are $50000$ odd numbers (extremes included). Thus, we have:

$\frac{50000a+1+3+5+\dotsb+99999}{1+3+5+\dotsb+99999}.$

Since the sum of the first $n$ odd numbers is $n^2$ , it follows that $1+3+5+\dotsb+99999 = (50000)^2.$

Now our expression becomes:

$\frac{50000a+(50000)^2}{(50000)^2}.$

Factor the numerator by 50000:

$\frac{50000(a+50000)}{(50000)^2} = \frac{a+50000}{50000}.$

Substitute back $a$ :

$\frac{100000+50000}{50000} = \frac{150000}{50000} = \boxed{3}.$

$x = \frac{100001+100003+100005+...+199999}{1+3+5+7+...+99999}$

Denominator is sum of odd numbers, so let's work that out first:

$\sum\limits_{n=0}^{N}(2n + 1)$

$\sum\limits_{n=0}^{N}{2n} + \sum\limits_{n=0}^{N}{1}$

$2\sum\limits_{n=0}^{N}{n} + (N+1)$

$\cancel{2}\frac{N(N+1)}{\cancel{2}} + (N+1)$

$N(N+1) + (N+1)$

$(N+1)(N+1)$

$(N+1)^2$

Cool! So the denominator is just $(N+1)^2$ . We'll work out the value of $N$ later.

Now, the numerator is a similar sum except the value $100000$ is added to each term:

$\sum\limits_{n=0}^{N}(2n + 1 + 100000)$

$\sum\limits_{n=0}^{N}(2n + 1) + \sum\limits_{n=0}^{N}100000$

$(N+1)^2 + (N+1)(100000)$

$(N+1)((N+1) + 100000)$

So the result of the whole series is just:

x = \frac{ \cancel{(N+1)}((N+1) + 100000)}{(N+1)^\cancel{2}}

$x = \frac{ (N+1) + 100000}{(N+1)}$

Now we need to work out the value of $N$ . The maximum odd number we compute is $99999$ , so:

$2N+1 = 99999$

$2N = 99998$

$N = 49999$

$\therefore x = \frac{ (N+1) + 100000}{(N+1)} = \frac{ 50000 + 100000}{50000} = \boxed{3}$

Writing large numbers is error prone so we can let $a = 100000$ Using formula $S = \frac{n}{2}(a+l)$ on both numerator and denominator.

$=\frac{\frac{1}{2}((a+1)+(2a-1))(a-1)}{\frac{1}{2}((1)+(a-1))(a-1)}$

simplify:

$=\frac{3a}{a}$

$=3$

$\frac{25 000(100 001+199 999)}{25 000(1+99 999)}$ = $\frac{300 000}{100 000)}$ = 3

the rule is =n(n+1)/2

so,100001+...................199999=n(n+1)/2=1.5 x 10^10

and 1+3+...........99999=n(n+1)/2=5 x 10^9

so,1.5 x 10^10/5 x 10^9=3

extracting 100000 from all of the terms in the numerator, and finding the number of terms , you can get (50000)(100000) + 1+3+5+7...99999. You might know that the sum of all odd numbers from 1 to n is ((n+1)/2)^2, applying that also to all of the original terms of the denominator, you get (50000)(100000) +(50000)(50000)/ (50000)(50000), canceling out 50000 from the numerator and denominator, after you combined (50000)(100000) and (50000)(50000) from the numerator, you get 150000/50000 and you get 3

Sum of odd numbers is n². So, between 1 .. 99999 has fifty thousands numbers, thus n² = 50000² = 2500000000.

100001 + 100003 + 100005 = (100000+1) + (100000+3) ... (100000+99999) (100000 * 50000) + (1+3+5 .... + 99999) = (100000 * 50000) + (50000²)

Thus (100000 * 50000) + (50000²) / (50000²) = 3

Pardon for my poor english :)

We can separate the big numbers in the numerator into 100000 +an odd number from 1 to 99999. We need to solve for the amount of positive odd numbers up to 99999. We can do this by using the generating function for odd numbers:

$a_n=2n-1$ Solving for 99999 as an odd number, we get the following:

$99999=2n-1$

$100000=2n$

$50000=n$ Now all we need to do is sum the top expression and the bottom expression as follows: $x=\frac{{100000*50000+\sum_{n=1}^{50000} 2n-1}}{\sum_{n=1}^{50000} 2n-1}$

Plugging this into a calculator with a computer algebra system, we get x=3.

The denominator and numerator each have $(99999-1) \div 2 + 1 = 50000$ numbers. Gauss Theorem : $\large{\frac {100001+100003+100005+ \cdots + 199999}{1+3+5+7+ \cdots + 99999} \newline = \frac{(100001+199999) \times \frac{50000} {2}}{(1+99999) \times \frac{50000}{2}} \newline = \frac{300000}{100000} \newline =} \LARGE{3}$

I used our good friend Gauss. Gauss' trick says that if there are an even number of terms, you can use this (for example) (I will use consecutive numbers starting from 1, other sequences work the same way): 1 + 2 + 3 + ... + n-2 + n-1 + n = $\\(\frac{n}{2}$ * (n + 1)) Then, we can apply it to the numerator (sum 300000 and 25000 terms), and the denominator (sum 100000 and also 25000 terms). We have the fraction $\frac{25000 * 300000}{25000 * 100000}$ , a very annoying way of writing 3 (divide the 25000s and 100000 off of both).

I got a simple method for the question. Add up the first number and the last number in the numerator, and then add up the first and the last number in the denominator. Divide the larger number by the smaller number. That's the correct answer!

set $1+3+5+7+...+99999=a$ , the numerator becomes $100000*(99999-1)/2+a$ , (there are $(99999-1)/2$ 100000s therefore the expression can become $100000*(99999-1)/2/a+1$ ; a=(first term + last term)*(number of terms)/2= $100000*(99999-1)/2/2$ so the expression becomes 2+1=3

Let's solve it in a very short time using the trick that the sum of the first n odd is n². Using this, The numerator is ((199999+1)/2)²-((100001-1)/2)² =(2 50000)²-50000² =3 50000². The denominator, ((99999+1)/2)²=50000 Solution is (3*50000²)/50000²=3

$\frac{(10^{5} \times 50000) + 50000^{2}}{50000^{2}} = \frac{50000^{2}(\frac{10^{5}}{50000} +1)}{50000^{2}} = \frac{10^{5}}{50000} +1 = \boxed{3}$

$\frac { (100,001+100,003+100,005+...+199,999) }{ 1+3+5+...+99,999 } =\quad \frac { (100,000+100,000+100,000+...+100,000)+(1+3+5+...+99,999) }{ 1+3+5+...+99,999 } \\ \quad \\ We\quad now\quad that\quad 1+3+5+...+99,999\quad is\quad the\quad sum\quad of\quad a\quad sequence\quad of\quad n\quad terms\quad where\quad the\quad nth\quad term\quad is\quad 2n-1.\quad We\quad want\quad to\quad know\quad the\quad number\\ of\quad terms\quad in\quad order\quad to\quad simplify\quad our\quad calculation.\quad \\ \\ If\quad the\quad last\quad term\quad of\quad the\quad sequence\quad is\quad 99,999=2n-1\quad \quad \rightarrow \quad n=50,000\\ \\ So\\ \frac { [(100,000+100,000+100,000+...+100,000)+(1+3+5+...+99,999)] }{ 1+3+5+...+99,999 } \\ \\ =\frac { 100,000*50,000+\sum _{ n=1 }^{ 50,000 }{ 2n-1 } }{ \sum _{ n=1 }^{ 50,000 }{ 2n-1 } } \\ =\frac { 100,000*50,000 }{ \sum _{ n=1 }^{ 50,000 }{ 2n-1 } } +\frac { \sum _{ n=1 }^{ 50,000 }{ 2n-1 } }{ \sum _{ n=1 }^{ 50,000 }{ 2n-1 } } \\ \\ =\frac { 5,000,000,000 }{ \sum _{ n=1 }^{ 50,000 }{ 2n } -\sum _{ n=1 }^{ 50,000 }{ 1 } } +1\\ \\ =\frac { 5,000,000,000 }{ 2\sum _{ n=1 }^{ 50,000 }{ n } -\sum _{ n=1 }^{ 50,000 }{ 1 } } +1\\ \\ =\frac { 5,000,000,000 }{ 2\frac { (50,000*(50,000+1) }{ 2 } -50,000 } +1\\ \\ =\frac { 5,000,000,000 }{ (50,000*50,001)-50,000 } +1\\ \\ =\frac { 5,000,000,000 }{ 50,000*50,001-50,000 } +1\\ \\ =\frac { 5,000,000,000 }{ 2,500,000,000 } +1\\ \\ =2+1\\ \\ =3\\ \\ \\$

There are 5000 terms in both the numerator and the denominator. Each of these pairs of terms can be added: Denominator: 1 + 99999 = 10000 2 + 99998 = 10000 3 + 99997 = 10000 ... ... 5001 + 4999 = 10000, Therefore the denominator equals 5000 x 10000 = 50,000,000 By parallel reasoning, the numerator reduces to 150,000,000 Therefore the fraction itself evaluates to 3.

This may also be solved by multiple cancellation.

The sum of $n$ sequential terms is $\frac{n}{2}(f+l)$ , where $f$ and $l$ are the first and last items respectively.

This gives us: $\frac{\frac{n}{2}(100001+199999)}{\frac{n}{2}(1+199999)}$ .

The $\frac{n}{2}$ cancels out, giving us: $\frac{100001+199999}{1+199999} = \frac{300000}{100000} = 3$ .

3 is the answer.

Sum of first $n$ odd integers is $n^2$ . So our denominator is $5000^2$ .

Numerator has $5000$ terms, each of which is $10000+i$ where the $i$ 's range through the first $5000$ odd integers. So the numerator is $10000\cdot 5000 + 5000^2 = 3*5000^2$ . Hence the overall value is $\frac{3*5000^2}{5000^2} = \boxed{3}$ .

1) Notice that the sum $1 + 2+ 3 + \dots + 99999$ contains $\frac{1+99999}{2} = 50000$ terms.

2) Thus both the denominator and the numerator consists of $n = 50000$ terms. The sum of the first $n$ odd integers are $n^2$ , and thus we have

$\dfrac{100001 + 100002 + \dots + 199999}{1+2+\dots+99999} = \dfrac{50000 \cdot 100000 + 50000^2}{50000^2}$

3) Carrying out the divison term-wise gives us $\frac{100000}{50000} + 1 = 2 + 1 = 3$ .

Sum of n consecutive odd numbers(starting from 1) give n^2 eg. 1+3=4=2^2 hence we can denote S=1+3+5.......99999 as 50000^2=25×10^8 so in numerator it is written (5×10^9)+S and denominator is S so by putting value of S it simplifies to 3

From the AP 1,3,5,7,9,11,13,15,17,19,21,..., start making ratios:

3/1=3; 1+3=4 see the little square size 2.

Increasing the denominator and numerator by one results in even sizes of squares.

And is always possible 2 green lines to divide every new square in 4 equal squares.

(5+7)/(1+3)=12/4=3; 4+12=16 see the larger square size 4.

And we can see every new square has 3 parts denominator and one part numerator.

(7+9+11)/(1+3+5)=27/9=3; 9+27=36 see the largest square size 6.

(9+11+13+15)/(1+3+5+7)=48/16=3; 16+48=64 and so on . . .size 8.

This way the ratio of 3 never changes.

. . . . . . . . . . . . . . . . . . . . . . . . . . . 75/25=3; 25+75=100 . . . size 10.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108/36=3; 36+108=144 . . . size 12.

. . . . . . . . . . . . . . . . . . . size 1000, size 1000000, size 1000000000 . . .

Multiply both numerator and denominator by 2 to make a copy in reverse order.

Numerator:

$100001+100003+100005+\ldots+199995+199997+199999$ $199999+199997+199995+\ldots+100005+100003+100001$

$300000+300000+300000+\ldots+300000+300000+300000$

Denominator:

$000001+000003+000005+\ldots+099995+099997+099999 \\ 099999+099997+099995+\ldots+000005+000003+000001$

$100000+100000+100000+\ldots+100000+100000+100000$

There is the same number of terms, so we get

$\frac{300000}{100000}=\boxed{3}$

The numerator and denominator is in the form of arithmetic progression. We need to get the sum of the progression on the numerator then divide it by the sum of the progression on the denominator.

S = n/2(a1 + an)

The answer is 3.

number of terms=50000 both in numerator & denominator Sum of Numerator = 100000 * 50000 + 50000 * 50000 (sum of 50000 odd numbers) = 5k *(1 lac+5k) Sum of Denominator = 5k * 5k (sum of 50000 odd numbers) Thus we have our final answer after cancelling out 5k both in numerator and denominator. Which is - 1,50,000/50,000 = 3

First step is to find out how many terms there are on the numerator and the denominator. It is quite obvious that there are the same number of terms in the numerator and the denominator.

Here, the last term of the numerator is 199999 ( $A_n$ ), first term is 100001 ( $a$ ), the common difference between two odd numbers is 2 ( $d$ ) $A_{ n }=a+(n-1)d\\ 199999=100001+2(n-1)\\ n=\frac { 199999-100001+2 }{ 2 } \\ n=50000$

Now we know,

$\underbrace { 1+3+5+7+..........+A_n } =n^{ 2 }$ $\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad n\quad terms$

So,

$1+3+5+7+...+99999=50000^{2}$

and,

$\quad 100001+100003+100005+100007+........+199999\\ =50000(100000)+50000^{ 2 }\\ =50000(100000+50000)\\ =50000\times 150000$

Thus we have,

$\large \quad \frac { 50000(150000) }{ 50000(50000) } \\ \large=\frac { 150000 }{ 50000 } \\ =\large \boxed { 3 }$

My solution was similar. $1+3+\cdots+(2n-1)=n^2\\ 1+3+\cdots+99999=n^2\Rightarrow\\ 2n-1=99999\Rightarrow n=50000\\ \text{similarly}\\ 1+3+\cdots+199999=k^2\Rightarrow k=100000.\\ \text{In the numerator we have}\\ 100001+100003+\cdots+199999=\\ (1+3+\cdots +199999)-(1+3+\cdots+99999)=k^2-n^2\\ \text{hence} \frac{100000^2-50000^2}{50000^2}=\frac {100000^2}{50000^2}-1=\left (\frac {100000}{50000}\right)^2-1=2^2-1=4-1=3$

Firstly: 100001+100003+...+199999 100000N+1+3+...+99999

Thus by breaking it apart between 100000N and 1 and dividing: 100000N/ (1+3+5+7+...+99999) +1 Equation (1)

So for the series expand the last 4 as well as the first 4 1+3+5+7+....+99993+99995+99997+99999

Then first and last sums up, the 2nd and 2nd last sums up etc. So that: 1+99999 + 3+99997 + 5+99995+.. =100000*N/2 Equation (2)

But the series can be shown to be showed by: N + (N-1)= 2N-1 Calculating the number of terms: 99999=2N-1 50000=N

Thus Equation (1) can be expressed as follows: 100000N/(100000*N/2) + 1 Equation (3)

By substituting 50000 into equation 3:

2+1 =3

My solution was this:

$\frac{100001 + 100003 + ... + 199999}{1+3+...+99999} = \frac{10000k + \sum\limits_{n=0}^{n=k-1}2n+1}{\sum\limits_{n=0}^{n=k-1}2n+1}=\frac{100000k}{\sum\limits_{n=0}^{n=k-1}2n+1}+1$

The sum equals:

$\sum\limits_{n=0}^{n=k-1}2n+1 = \frac{1+2(k-1)+1}{2}k = k^2$

And the result is:

$\frac{100000}{k}+1 = 3$

Because $k = \frac{100000}{2}$

This is simple AP nothing else

if you reduce this down to 101-199 over 1+3 ... 99 you get n^2+100n/n^2 ==> there are 50 terms between => n = 50..

50^2+100(50)/50^2
(2500+5000)/2500 7500/2500 3

Sum of n odd terms = ${n}^{2}$

n =( last term +1) / 2

--> n=50000.

Now express all numbers in terms of 100000 and take it common.

100000+3,10000+5,.....,100000+99999 $=\frac{(100000×n) + 1+3+5+7....+99999}{1+3+5+7+...+99999}$ $=\frac { (100000×n )+(n^2) } { (n^2)}$ $=\frac{100000}{n}+1$ ; n=50000 $= 2+1 = 3$

$\begin{aligned} \frac{100001+100003+100005+ \ldots + 199999}{1+3+5+\ldots+99999} &= \frac{100000+\ldots+100000+1+3+5+\ldots+99999}{1+3+5+\ldots+99999} \\ &=\frac{50000\cdot100000+1+3+5+\ldots+99999}{1+3+5+\ldots+99999} \\ &=\frac{50000\cdot100000}{50000^{2}}+1 \\ &=2+1 \\ &=3 \end{aligned}$

Both Nr. & dr are Aps, No. of termsof Nr= (199999 -100001)/2 + 1=49999, no. of terms in Dr. also 49999, Then, Sum of Nr / Dr =n/2(a +an) divided by n/2(b+bn) = 49999/2(100001+199999) by49999/2x(1+99999) = 150000/50000 =3

Numerator can be totalled by adding 2 pairs --> 100001 + 199999, 100003 + 199997 and so on for 50000 times ==> 300000 * 50000 Denominator is similarly --> 1+99999, 3+99997 and so on for 50000 times ==> 100000 * 50000 Numerator / Denominator = 300000/100000 = 3!

Numerator:

First, pair up the first and last terms, then the second and second-to-last terms, and so on.

100,001 + 199,999 = 300,000, and 100,003 + 199,997 = 300,000, etc.

From 100,001 to 200,000 (counting the even numbers), there are 100,000 numbers, and therefore 50,000 pairs that add up to 300 ,000. Since we are dealing with the sum of the odd pairs only, there are 25 ,00 pairs that add up to 300,000. For simplicity, we will not evaluate this yet. Instead, the sum of the numerator is

300,000 x 25,000

Denominator:

Using the same idea:

1 + 99 999 = 100 000, 3 + 99 997 = 100 000, etc.

From 1 to 100,000, there are 100,000 numbers, or 50,000 pairs that add up to 100,000. There are 25,000 odd-number pairs. The sum is

100,000 x 25,000

Final Sum:

=(300,000 x 25,000) / (100,000 x 25,000)

=(300,000) / (100,000 ), where the 25,000s cancel out.

= 3

(100000+1+100000+3+...)/1+3+5.... A total of 50000 nos so (100000*50000 + 50000^2)/50000^2 = 3 (since 1+3+5+...n= n^2)

for denominator part summation of no.s is (99999 100000/2)-49999 50000=2500000000, for numerator we can write 100000+1+100000+2+100000+3+.........+100000+99999= 5000000000+2500000000=7500000000, 7500000000/2500000000=3

Finding the two sums and then dividing is the hard way. The average term on top is the average of $100,001$ and $199,999$ ; thus it is $150,000$ . The average on the bottom is similarly found; it is $50,000$ . And the top and the bottom have equally many terms. So it's just $150,000/50,000$ , which is $3$ .

I'd have written $1+3+5+7+9+11+\cdots+99999$ (with \cdots) or at least $1+3+5+7+9+11+\ldots+99999$ rather than $1+3+5+7+9+11+...+99999$ (with just three dots: ... ). Notice the resulting asymmetry: the leftmost dot is closer to the nearby plus sign than the rightmost dot is to its nearby plus sign. And if you do this is genuine LaTeX (which is not limited to math notation) as opposed to this stripped-down thing, the use of actual dots will neglect proper spacing between the dots.