Square Summation

This solution uses Partial Fractions - Linear Factors

Let $S$ be the summation, therefore,

$\begin{aligned} S & = \frac{3^2+1}{3^2-1} + \frac{5^2+1}{5^2-1} + \frac{7^2+1}{7^2-1} + ... + \frac{99^2+1}{99^2-1} \\ & = \sum_{k=1}^{49} \frac{(2k+1)^2+1}{(2k+1)^2-1} \hspace{.4cm} \text{ (Writing in summation notation)} \\ & = \sum_{k=1}^{49} \left( 1 + \frac{2}{(2k+1)^2-1} \right) \\ & = \sum_{k=1}^{49} \left( 1 + \frac{2}{[(2k+1)-1][(2k+1)+1]} \right) \\ & = \sum_{k=1}^{49} \left( 1 + \frac{2}{4k(k+1)} \right) \\ & = \sum_{k=1}^{49} \left( 1 + \frac{1}{2}\left[\frac{1}{k} - \frac{1}{k+1} \right] \right) \hspace{.4cm} \text{ (Using Partial Fractions)} \\ & = \sum_{k=1}^{49} 1 + \frac{1}{2} \left( \sum_{k=1}^{49} \frac{1}{k} - \sum_{k=2}^{50} \frac{1}{k} \right) \\ & = 49 + \frac{1}{2} \left( \frac{1}{1} - \frac{1}{50} \right) \\ & = \boxed{49.49} \end{aligned}$

(a^2 + 1)/(a^2 - 1) = 1 + 2/(a^2 - 1) = 1 + 1/(a - 1) - 1/(a + 1), so that means we can write the sum as

(1 + 1/2 - 1/4) + (1 + 1/4 - 1/6) + (1 + 1/6 - 1/8) + ... + (1 + 1/98 - 1/100)

= 49x1 + 1/2 - 1/100

= (4900 + 50 - 1)/100

= 4949/100

I let $x=4$ , so the expression becomes;

$\frac{x+1}{x}+\frac{3x+1}{3x}+\frac{6x+1}{6x}+.........+\frac{4950x+1}{4950x}$

( I figured out the denominator by finite difference sequence)

Simplify the expression as;

$(1+\frac{1}{x})+(1+\frac{1}{3x})+(1+\frac{1}{6x})+............(1+\frac{1}{4950x})$

Group it:

$(1+1+1+1+...........1)+(\frac{1}{x}+\frac{1}{3x}+\frac{1}{6x}+..............\frac{1}{4950x})$

Substitute $x=4$ ;

$(1+1+1+1+............1)+(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+..............\frac{1}{19800})$

Here's where the telescopic sequences comes in;

$\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+............\frac{1}{19800}=(\frac{1}{2}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{6}-\frac{1}{8})+..............+(\frac{1}{198}-\frac{1}{200})$

All the middle terms cancel out leaving:

$\frac{1}{2}-\frac{1}{200}=0.495$

Now the 1's. There are 49 1's in the expression above, so the answer is $49(1)+0.495=\boxed{49.495}$

The sum can be written as $S=\sum_{n=1}^{49}{(2n+1)^2+1 \over (2n+1)^2-1} =\sum_{n=1}^{49}{(2n+1)^2+1 \over (2n+2)(2n)} =\frac{1}{4}\sum_{n=1}^{49}{(2n+1)^2+1 \over n(n+1)}$

Now consider the term

${(2n+1)^2+1 \over n(n+1)} = {(n+(n+1))^2+1 \over n(n+1)} ={(n^2+(n+1)^2 +2n(n+1) + 1 \over n(n+1)}$

$=\frac{n}{n+1}+\frac{n+1}{n}+2 +\frac{1}{n(n+1)} =4+2\left(\frac{1}{n}-\frac{1}{n+1}\right)$ Using partial fractions to simplify. Now its straight forward

$S=\frac{1}{4}\sum_{n=1}^{49} \left( 4+2\left(\frac{1}{n}-\frac{1}{n+1}\right) \right) =\sum_{n=1}^{49}1 +\frac 12 \sum_{n=1}^{49} \left(\frac{1}{n}-\frac{1}{n+1}\right) =49+\frac12 (1-\frac{1}{50})={\bf 49.49}$

simple solution, (10/8)+(26/24)+(50/48)+--------------(9802/9800); = 1.25+1.0833+1.0416---------------+1.0002; here we have 49 terms ,so ans should be [49+ something]. we can easily observe that numbers after decimal are conversing to zero, hence (0.25+0.0833+0.0416+----------+0.0002)= 1 (approx). hence correct ans =50(approx).

use the Sigma Sum from 1 to 49 of [(2n+1)^2+1]/[(2n+1)^2-1] = correct answer is 49.49,

Write a solution. 50.33 is an approximate figure. 49.49 indicated by Mr. Cheong seems to be way out!