A Long Sum

It would be laborious for us to calculate the sum, so instead of that, let us try to trace the pattern.

$\frac{3}{1^{2}2^{2}}$ $=$ $\frac{3}{4}$

$\frac{3}{1^{2}2^{2}} + \frac{5}{2^{2}3^{2}}$ $=$ $\frac{8}{9}$

$\frac{3}{1^{2}2^{2}} + \frac{5}{2^{2}3^{2}} + \frac{7}{3^{2}4^{2}}$ $=$ $\frac{15}{16}$

So, now, we can generalize the sums above as follows.

$\frac{3}{1^{2}2^{2}} + \frac{5}{2^{2}3^{2}} + \frac{7}{3^{2}4^{2}} + ... + \frac{2n+1}{n^{2}(n+1)^{2}}$ $=$ $\frac{(n+1)^{2} - 1}{(n+1)^{2}}$

Therefore,

$\frac{3}{1^{2}2^{2}} + \frac{5}{2^{2}3^{2}} + \frac{7}{3^{2}4^{2}} + ... + \frac{41}{20^{2}21^{2}}$ $=$ $\frac{21^{2} - 1}{21^{2}}$ $=$ $\frac{440}{441}$

Comparing the fraction with $\frac{a}{b}$ , we get $a = 440$ and $b = 441$ Thus, $a+b = \boxed {881}$

My solution to solve this problem is "pattern recognition". So, this my steps:

1. First, see the patern by adding one by one of the sequence above (sum n -th term of sequence). Initialize that sum of 1 -st term is $\frac{3}{4}$ , then the sum 2 -nd term is you add 1 -st and 2 -nd term of sequence first, that are $\frac{3}{1^{1}2^{2}}$ and $\frac{5}{2^{2}3^{2}}$ , sum must be $\frac{8}{9}$ simplified from $\frac{32}{36}$ .

2. Next, you can continue the summing of the 3 -rd term, sum of $\frac{3}{1^{1}2^{2}}$ , $\frac{5}{2^{2}3^{2}}$ , and $\frac{7}{3^{2}4^{2}}$ . So, sum is $\frac{15}{16}$ , the simplified from $\frac{45}{48}$ . Continue, your addtion, and you can see the pattern below.

3. The patter of sum n -th term of sequence are $\frac{3}{4}$ , $\frac{8}{9}$ , $\frac{15}{16}$ , $............$ . So, you can see the patern now, that the sum of sequence above is the sum until 40 -th term. The pattern are the denominator (the square of $n+1$ , $*n*$ from $1, 2, ....$ ) and the numenator (sum of odd number from $3$ to $41$ , you can use arithmetic progression).

4. Finally, sum is $\frac{sum of odd number}{(n+1)^2}$ = $frac{440}{441}$ . $a$ is 440 and $b$ 441, then $a + b$ is $400 + 441$ = $\boxed{881}$ .

Since the nth term of the equation can be written as

$\frac{1+2r}{r^{2}\times({1+r}^{2})} =\frac{1}{r^{2}}-\frac{1}{(1+r)^{2}}$

putting r=1,2,3,4.....20

$\frac{1}{1^{2}}-\frac{1}{2^{2}}+\frac{1}{2^{2}}-\frac{1}{3^{2}}+...........+ \frac{1}{19^{2}}-\frac{1}{20^{2}}+\frac{1}{20^{2}}-\frac{1}{21^{2}}$

Common term will cancel

so we have $1-\frac{1}{21^{2}}=\frac{441-1}{441}=\frac{440}{441}$

$= \boxed{881}$

Note: This solution is similar to the ones posted below, but I still decided to write it because I felt it would be much clearer in LaTeX. I would be inserting motivations within the solution itself in [ ] brackets.

Motivation 1 As the series is named Telescoping, we ought to look out for patterns (cancellation of terms) in this question.

Solution

Observe that all terms in the question can be expressed as $\dfrac{2x+1}{x^2(x+1)^2}$ . Expanding this [why? because we are looking for patterns or way to simply the expression, so expansion might do the trick], we get $\dfrac{2x+1}{x^2\cdot (x^2 +2x +1)}$ .

[Notice that we have 2 similar expression in the denominator and numerator of the fraction. This expression is simply $2x+1$ .] So, how can we construct variable(s) such that we obtain a telescoping sequence? Here comes the trick! :) Simply add and subtract $x^2$ to the numerator to obtain $\dfrac{(2x+1+x^2)-x^2}{x^2(x^2+2x+1)}$ . After doing this, note that we can simplify the expression further by cancelling the common terms top and bottom of the fraction! We get, $\dfrac{(2x+1+x^2)-x^2}{x^2(x^2+2x+1)} =\dfrac{1}{x^2} - \dfrac{1}{(x^2+2x+1)}$

Recalling $x^2 +2x+1 = (x+1)^2$ , this simplifies further into $\dfrac{1}{x^2} - \dfrac{1}{(x+1)^2}$ .

Therefore, fitting this back into our question we get, $\frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{2^2} - \frac{1}{3^2} + ... + \frac {1}{20^2} - \frac{1}{21^2}$ . All terms cancel out to obtain: $\frac{1}{1^2} -\frac{1}{21^2} = \boxed{\frac{440}{441}}$ . We are done :)

First, recognize the pattern of the sum and you will get $\sum_{n=1}^{20} \frac{2n+1}{n^2(n+1)^2}$ This can be decomposed using partial fractions. After doing this you should have $\sum_{n=1}^{20} \frac{1}{n^2} - \frac{1}{(n+1)^2}$ Then plug in $n=1, 2, 3, ..., 20$ We end up with $(1- \frac{1}{4}) + (\frac{1}{4} - \frac{1}{9}) + (\frac{1}{9} - \frac{1}{16}) + ... + (\frac{1}{20^2} - \frac{1}{21^2})$ After cancelling, we are left with $1 - \frac{1}{21^2} = \frac{440}{441}$ So, the final answer is $440 + 441 = 881$

The whole expression can be expressed as:

$\sum_{i=1} \frac {a+ (a+1)} {a^2 (a+1)^2}$

where n = 20.

Simplifying the expression,

It becomes $\sum_{i=1} \frac {2a + 1} {a^2 (a+1)^2}$

Using partial fractions to allow to telescope the series:

$\frac {2a+1}{(a^2)(a+1)^2} = \frac {A}{a^2} + \frac {B}{a+1} + \frac {C}{(a+1)^2}$

$\frac {2a+1}{(a^2)(a+1)^2} = \frac {A(a+1)^2 + B[a(a+1)] + Ca^2}{a^2(a+1)^2}$

$2a + 1 = A(a+1)^2 + B[a(a+1)] + Ca^2$

If we let a = 0, A = 1; if we let a = -1, C = -1

Having obtained A and C, we will get B, and B = 0.

The partial fraction of $\frac {2a+1}{a^2(a+1)^2}$

is

$\frac {2a+1}{a^2(a+1)^2} = \frac {1}{a^2} + 0 + \frac {-1}{(a+1)^2}$

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

As this allows us to telescope the series, for n = 20

$\sum_{i=1} \frac {1}{a^2} - \frac {1}{(a+1)^2} = [1 - \frac {1}{(a+1)^2} ] = 1 - \frac {1}{441} = \frac {441}{441} - \frac {1}{441} = \frac {440}{441}$

If a = 440, and b = 441:

$440 + 441 = \boxed {881}$

$\sum_{n=1}^{20}\frac{2n+1}{(n^2)(n+1)^2}=\sum_{n=1}^{20}\frac{1}{n^2}-\sum_{n=1}^{20}\frac{1}{(n+1)^2}=\sum_{n=1}^{20}\frac{1}{n^2}-\sum_{n=2}^{21}\frac{1}{(n^2)}=1-\frac{1}{21^2}=\frac{440}{441}=\frac{A}{B}$ $\rightarrow A+B=\color{#3D99F6}{\boxed{881}}$

i just try to sum 3/2^2+5/2^2*3^2 and i get 8/9

then i try 3/2^2+5/2^2 3^2+7/3^2 4^2 and i get 15/16

then i try 3/2^2+5/2^2 3^2+7/3^2 4^2+9/4^2*5^2 and i get 24/25 until i figure out it is strange enough

i think the conclusion is (n^2-1)/n^2 with n is the largest denominator ex: 7/3^2*4^2 >>>n is 4

then 41/(20^2*21^2) n is 21 then the result is 21^2-1/21^2=440/441=a/b

a+b=440+441=881

A Classical case OF " TELESCOPIC SERIES" FIRST, WE NEED TO FIND OUT Tn.. Tn=(2n+1)/{(n^2)*(n+1)^2}.... on rearranging the Tn can be presented as ..... Tn= 1/n^2 - 1/(n+1)^2 now put n=1.... we get T(1)=1 - 1/4 now put n=2.... we get T(2)= 1/4 - 1/9
.......and so on till T(20)= 1/400 - 1/441 Now add the following and notice that the terms get cancelled in a TELESCOPIC WAY ..(hence the name).. finally you get....the sum as 1 - 1/441 = 440/441 ...(a/b) a+b = 440 + 441 = 881.....ANSWER...

the problem can be stated as $=\frac{ 3 }{ 1*4 }+\frac{ 5 }{ 4*9 }+.....+\frac{ 41 }{ 441 }$ $\frac{ 4-1 }{ 1*4 }+\frac{ 9-4 }{ 4*9 }+.....+\frac{ 441-400 }{ 41 }$ $1-\frac{ 1 }{ 4 }+\frac{ 1 }{ 4 }+\frac{ 1 }{ 9 }-\frac{ 1 }{ 9 }-.....+\frac{ 1 }{ 400 }-\frac{ 1 }{ 441 }$ $1-\frac{ 1 }{ 441 }=\frac{ 440 }{ 441 }$ So a+b=841

Since this sequence look kind of complicated, let's look term by term.

First, we look at the first fraction $\frac{3}{1^12^2}$ . This is equal to $\frac{3}{4}$

Next, we look at the sum of the first two fractions, $\frac{5}{2^2 3^2} + \frac{3}{4}= \frac{5}{36}+\frac{3}{4} = \frac{8}{9}$

The sum of the first three fractions is $\frac{15}{16}$ , and so on.

We can continue calculating some more.

The sum of the first $n$ terms of this sequence is, for $n = 1, 2, 3, 4, \ldots$ ,

$\frac{3}{4}, \frac{8}{9}, \frac{15}{16}, \frac{24}{25}, \frac{35}{36}, \ldots$ .

We notice two things about this sequence.

1) The numerator is always one less than the denominator. 2) The denominator is always a perfect square and it is increasing by 1 perfect square each time - 4, 9, 16, 25...

Since we want the sum of the first twenty terms, our denominator should be $(20+1)^2 = 441$ following the sequence. Our numerator should, therefore, be $441-1 = 440$

Adding, we have $441+440 = \boxed{881}$

The general term is 2x+1/x^2(x+1)^2

which can be written as 2x+1+x^2-x^2/x^2(x^2+2x+1)

so we got 1/x^2-1/x^2+2x+1 =1/x^2-1/(x+1)^2

After we cancelled the term we got 1/1^2-1/21^2

which is equal to 440/441 a+b=881

First, you need to find a pattern by adding in this formation.

$\frac {3}{1^1 * 2^2}=\frac {3}{4}$

$\frac {3}{4}$ + $\frac {5}{2^2*3^2}=\frac {8}{9}$

$\frac {8}{9}$ + $\frac {7}{3^2*4^2}=\frac {15}{16}$

As you can see, the pattern for the denominators is $(t+1)^2$ where $t$ is the term number. The numerator is found by subtracting one from the denominator. Since 41 is the 20th term, the sum is $\frac {440}{441}$ , which is co-prime. $440 + 441=881$ .

all the fractional numbers of the series can be generalized to a^2 - b^2/ (a^2 . b^2 )..... is anything else required , euler .... ;)

You can see that the term of this sum can be write like that: (2n+1)/[n^2(n+1)^2]
It´s a good idea to separate this to: (2n+1)/[n^2(n+1)^2] = a/n^2 + b/(n+1)^2 , doing the equation you can find that a=1 and b=-1. So you have sum of (1/n^2 - 1/(n+1)^2) of n=1 to n=20 doing that you have 440/441, and 440+441= 881

The general term is 2x+1/x^2(x+1)^2 which can be written as 2x+1+x^2-x^2/x^2(x^2+2x+1) so we got 1/x^2-1/x^2+2x+1 =1/x^2-1/(x+1)^2 After we cancelled the term we got 1/1^2-1/21^2 which is equal to 440/441 a+b=881

[3/1^2 2^2]+[5/2^2 3^2]+[7/3^2 4^2]+---------------------------+[41/20^2 21^2]=

3 can be written as (4-1),similarly 5=9-4, 7=16-9 respectively.

[(4-1)/4 1]+[(9-4)/9 4]+[(16-4)/16 9]+---------------------------+[(441-400)/441 400]=

1-(1/4)+(1/4)-(1/9)+(1/9)-(1/16)+(1/16)+----------------------(1/400)+(1/400)-(1/441)=

common terms cancels out & we left with

1-(1/441)=(441-1)/441=440/441=a/b

a+b=440+441=881, is the required answer.

Sorry I did not see Michael David Sy solution befor. Mine is the same method put with displaystyle. The solution by Happy Melodies is really nice and shows his intelligence. How ever I thought of giving a solution that can apply to Telescopic series in general. It is known as splitting into partial fractions.
$\displaystyle\ For~this~problem ~the~ nth~ term~is:-~\frac{2n + 1} { n^2*( n + 1 )^2 }$
$\displaystyle\ So :-~\frac{2n + 1} { n^2*( n + 1 )^2 }=\frac{A}{n}+\frac{B}{n^{2}}+ \frac{C}{n+1} + \frac{D}{(n+1)^2}$ $\displaystyle\ where~we~need~to~find~out~ the~ values~ of~ rationals~ A,~ B, ~C,~ D.$
$\displaystyle\ ~Multiplying~ both~ sides~ by~ n^2*( n + 1 )^2~~we~get~~~~~~~~~$

$\displaystyle\ 2n + 1 = A{ n*( n + 1 )^2} +B{ ( n + 1 )^2} + C{ n^2*( n + 1 ) } + Dn^2$

$\displaystyle\ Equating~ n^3,~ n^2,~ n ,~ and~ constant~ terms~ of~ both~ the~ sides,~ we~ get :-$
$\displaystyle\ 0 = (A + C)* n^3, 0 = (2A + B + C + D)* n^2, 2n = (A + 2B)*n , 1 = B.$

$\displaystyle\ gives~ A+C=0,~~2A+B+C+D=0,~~A+2B=2,~~~~~$ $\displaystyle\ B=1.~~A=0,~~A=-C=0,~~0+1+0+D=0-->D=-1$

$\displaystyle \frac{2n + 1}{n^2*( n + 1 )^2} = \frac{1}{n^2} - \frac{1}{( n + 1 )^2}~~these~are~two~partial ~ fractions.$

$\displaystyle \sum_{n=1}^{21} \frac{1}{n^2 } - \sum_{n=1}^{21} \frac{1}{(n+1)^2 }$
$\displaystyle =1+\sum_{n=2}^{21} \frac{1}{n^2 } - \sum_{n=1}^{20} \frac{1}{(n+1)^2 }- \frac{1}{21^2.}~~~~~~~~~~~But~\sum_{n=2}^{21} \frac{1}{n^2 } = \sum_{n=1}^{20} \frac{1}{(n+1)^2 }$
$\displaystyle =1+[0]- \frac{1}{21^2} = 1 - \frac{1}{441} = \frac{440}{441} = \frac{a}{b}$
$\displaystyle a+b=440+441 =\boxed{881}$

If u remember, in your school days, you must have made the observation that: (note that * means 'squared') 1+3 = 2* 1+3+5=3* 1+3+5+7=4* and so on.. So, 5 = 3 -2 7=4 -3

So the first term is = (2 -1 )/1 2 =1/1* - 1/2* the second term = 1/2* - 1/3*

So consecutive terms cancel and finally we're left with 1- 1/21*=440/441 Hence, 881 is the answer

if you observe the pattern, it is of the form (i + (i+1))/(i^2)(i+1)^2. then we end up at the expression (2i+1)/(i(i+1))^2. then we add and subtract i^2 in the numerator. thus we can simplify the numerator to (i+1)^2 - i^2. thus we can seperate the fraction into, (1/i^2) - (1/(i+1)^2).

for i=1: (1/1^2) - (1/2^2) for i=2: (1/2^2) - (1/3^2) .... .. .. . . for i=20: (1/20^2) - (1/21^2)

adding all of which we end up at : (1/1^2) - (1/21^2) = ((21^2 - 1)/21^2) = 440/441

a=440 b=441 a+b = 881 :)

Since $\frac{2k+1}{k^2(k+1)^2}=\frac1{k^2}-\frac1{(k+1)^2}$ , the sum telescopes to $\frac1{1^2}-\frac1{21^2}=\frac{440}{441}$

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