Question

Evaluate


`\rm (3^2+1)/(3^2-1)+(5^2+1)/(5^2-1)+(7^2+1)/(7^2-1)+\ldots+(101^2+1)/(101^2-1) =`
With step by step explanation !​

Answer 1

It's easier to deal with the symbolic sum (in sigma notation),

`\displaystyle\sum_(k=1)^(50)((2k+1)^2+1)/((2k+1)^2-1)`

Expanding the terms in the fraction, computing the quotient, and decomposing into partial fractions gives

`((2k+1)^2+1)/((2k+1)^2-1) = (4k^2 + 4k + 2)/(4k^2 + 4k)`

`=\frac12*(2k^2 + 2k + 1)/(k^2 + k)`

`=\frac12\left(2+\frac1{k(k+1)}\right)`

`=\frac12\left(2 + \frac1k - \frac1{k+1}\right)`

and it's the latter two terms that reveal a telescoping pattern.

In case you need more details about the partial fraction decomposition, we are looking for coefficients a and b such that

`\frac1{k(k+1)}=\frac ak+\frac b{k+1}`

or

`1 = a(k+1) +bk =(a+b)k+a`

which gives a = 1, and a + b = 0 so that b = -1.

Our sum has been rearranged as

`\displaystyle\frac12\sum_(k=1)^(50)\left(2+\frac1k-\frac1{k+1}\right)=\sum_(k=1)^(50)1+\frac12\sum_(k=1)^(50)\left(\frac1k-\frac1{k+1}\right)=50+\frac12\sum_(k=1)^(50)\left(\frac1k-\frac1{k+1}\right)`

The remaining telescoping sum is

1/2 [(1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + … + (1/48 - 1/49) + (1/49 - 1/50) + (1/50 - 1/51)]

and you can see how there are pairs of numbers that cancel, so that the sum reduces to

1/2 [1/1 - 1/51] = 1/2 [1 - 1/51] = 1/2 × 50/51 = 25/51

So, our original sum ends up being

`\displaystyle\sum_(k=1)^(50)((2k+1)^2+1)/((2k+1)^2-1) = 50 + (25)/(51) = \boxed{(2575)/(51)}`