Let for \(\mathrm{n}=1,2, \ldots, 50, \mathrm{~S}_{\mathrm{n}}\) be the sum of the infinite geometric progression whose first term is \(\mathrm{n}^{2}\) and whose common ratio is \(\frac{1}{(\mathrm{n}+1)^{2}}\). Then the value of \(\frac{1}{26}+\sum_{n=1}^{50}\left(S_{n}+\frac{2}{n+1}-n-1\right)\) is equal to

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Sequences and Series

Let for

n = 1, 2, \dots, 50, S_{n}

be the sum of the infinite geometric progression whose first term is

n^{2}

and whose common ratio is

\frac{1}{( n + 1 ) ^{2}}

. Then the value of

\frac{1}{26} + \sum_{n = 1}^{50} (S_{n} + \frac{2}{n + 1} - n - 1)

is equal to

Answer: 41651

For an infinite GP with first term

a

and common ratio

r

(where

∣ r ∣ < 1

), the sum is

S = \frac{a}{1 - r}

.

Here,

S_{n} = \frac{n ^{2}}{1 - \frac{1}{( n + 1 ) ^{2}}} = \frac{n ^{2} ( n + 1 ) ^{2}}{( n + 1 ) ^{2} - 1} = \frac{n ^{2} ( n + 1 ) ^{2}}{n ( n + 2 )} = \frac{n ( n + 1 ) ^{2}}{n + 2}

.

Expanding:

S_{n} = \frac{n ( n ^{2} + 2 n + 1 )}{n + 2} = \frac{n ^{3} + 2 n ^{2} + n}{n + 2} = n^{2} + 1 - \frac{2}{n + 2}

.

Therefore:

S_{n} + \frac{2}{n + 1} - n - 1 = n^{2} - n + \frac{2}{n + 1} - \frac{2}{n + 2}

.

Summing from

n = 1

50

\sum_{n = 1}^{50} (n^{2} - n) + 2 \sum_{n = 1}^{50} (\frac{1}{n + 1} - \frac{1}{n + 2})

.

The first sum is

\frac{50 \cdot 51 \cdot 101}{6} - \frac{50 \cdot 51}{2}

.

The second sum telescopes:

2 (\frac{1}{2} - \frac{1}{52}) = 1 - \frac{1}{26}

.

Adding

\frac{1}{26}

gives the final answer: 41651.

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Integer answer