A cute odd triangle Part 1

We are trying to find the sum of first 100 odd integers i.e, 1,3,5,.....199.

Split the sum into the following pairs,

$1+199 = 200$

$3 + 197 = 200$

$5 + 195 = 200$ . . . .

. . . . . $99 + 101 = 200$

We find that each of these have sum equal to $200.$

And there are $50$ pairs of numbers. So the sum is $50\times200 = \boxed{10,000}$

$\displaystyle\sum_{r=1}^{n} 2r - 1 = 2 \displaystyle\sum_{r=1}^{n} r - \displaystyle\sum_{r=1}^{n} 1 = n(n+1) - n = n^2$ $\therefore\displaystyle\sum_{r=1}^{100} 2r - 1 = 100^2 = 10 000$

Since sum to n terms is given by

S=n/2 (2a+(n-1)d)

Where,n is the number of terms,a is the first term and d is the common difference

Since for first row,n=1 and for 2nd row n=2

Therefore for 100th row,n=100,a=1,d=2

S=100/2 (2 (1)+(100-1)2)

S=10000

We note that:

$\begin{array} {cll} 1 & = 1 & = 1^2 \\ 1+3 & = 4 & = 2^2 \\ 1+3+5 & = 9 & = 3^2 \\ 1+3+5+7 & = 16 & = 4^2 \\ 1+3+5+7+9 & = 25 & = 5^2 \\ 1+3+5+7+9+ 11 & = 36 & = 6^2 \\ ... & ... & ...\\ \Rightarrow 1+3+5+7+9+...+ (2n-1) & & = n^2 \\ \Rightarrow 1+3+5+7+9+...+ 199 & = 100^2 & = 10000 \end{array}$

The first row contains one odd number, the second row contains two odd numbers, the third row contains three odd numbers..........

Similarly the hundredth row will contain 100 odd numbers.

Since the difference between every odd number is 2, we can see that the sequence of odd numbers is an AP, and can figure out the sum of $n$ odd numbers, which is $n^2$ which here gives us $10000$

second row's last digit=2+(2-1)=3; third row's last digit=3+(3-1); hundredth row's last digit=100+(100-1)=100+99=199 ;; 1+3+5+7+.............+199=(1+2+3+4+.......+200)-(2+4+6+8+....+200) =20100-10100=10000

Think simple... We know that the sum of n odd numbers results in its square. So the sum of 100 odd numbers is equal to 100^2 = 10000

1st row = 1^2 2nd row = 2^2

So on...

so 100th row = 100^2 =10000

$1 = 1 = 1^{2}$

$1 + 3 = 4 = 2^{2}$

Using mathematical induction, let $\sum_{i=1}^k (2i - 1) = k^{2}$ be the induction hypothesis.

And then, we solve for $n = k + 1$ :

$\sum_{i=1}^{k + 1} (2i - 1) =$

$\sum_{i=1}^k (2i - 1) + (2 \times (k + 1) - 1) =$

$\sum_{i=1}^k (2i - 1) + (2k + 1) =$ (here we use the induction hypothesis)

$k^{2} + 2k + 1 =$

$(k + 1)^{2}$

Therefore, the summation of the first $n$ odd integers can be represented as $n^2$ . So, the sum of all numbers in the 100th row is $100^2 = 10000$ .

Since the difference of the sums of the rows is a linear pattern (1,3,5,7...), we know that this is a quadratic equation. Since the equation of the difference is d=2x-1 (where x is the row number), we know that a in ax^2+bx+c must be 2/2 (because in the format mx+b, m is 2), or 1. If we go back and deduce the fact that the sum of the 0th row is 0 (because there are no numbers there!), we know that c must be 0. Plugging in the point (1,1), and solving for b will yield b=0. So the final equation is x^2. Now since we're asked to find the sum of all the numbers in the 100th row, we can simply perform 100^2, and get 10000.

sum of first n odd numbers = n^2 and now , sum of first 100 odd numbers = 100^2 = 10000

in class 8th in our school we learned a method of finding squares by using odd numbers.

the method is:

for finding n ^{2} we just need to find the (sum of first n odd numbers). we can find a similar question here ;

so we just need to find n ^{2} to get the sum of n odd numbers

i.e.,100 ^{2}=10000 which is our answer. you can try this method for other numbers also for verification . Chung Kevin, a very good question.

On the 100th row there will be all odd numbers under 200, Each odd number has a corresponding number which when added together = 200 For example 1+199 or 3+197 Because there are 50 such pairs. The answer = 200 * 50 = 10000

See the pattern

n= 1 -----> 1 =1^2; n= 2 -----> 4 =2^2; n= 3. -----> 9 =3^2; n= 4. -----> 16 =4^2; . . . n = =======> n = n^2; so, <== > n=100 ------> 100^2 =10.000

The sum of an arithmetic series is:

S= (n/2)(a1-an)

Where n represents the number of terms.

The 100th odd number is 199.

Therefore, we have:

S = (100/2)(1+199)=(50)(200)=10000

we need to compute sum (2n-1) ; n=1 to 100 --- Odd numbers

= {sum (n); n=1 to 199 } - {sum(2n); n=1 to 99} --- sum of series minus sum of even numbers

= 199 200/2 - 2 99 100/2 = (199-99) 100 = 10000

The sum of the numbers in n th row = n^2

So the answer = 100^2 = 10000

I noticed that the sum of each row n corresponds to the square of n . Therefore the sum of the numbers in the 100th row would be the square of 100 which is 10000.