Let a1=1 and for n 1, an+1 = 12an + n^2-2n-12n^2(n+1)^2. Then n=1^ (an - 2n^2) is equal to:

Given: a1=1 and for n 1, an+1 = 12an + n^2-2n-12n^2(n+1)^2 Find: n=1^ (an - 2n^2) Define a new sequence bn = an - 1n^2 Then an = bn + 1n^2, an+1 = bn+1 + 1(n+1)^2 Substituting into the recurrence, bn+1 + 1(n+1)^2 = 12 (bn + 1n^2) + n^2-2n-12n^2(n+1)^2 So, bn+1 = 12bn + 12n^2 + n^2-2n-12n^2(n+1)^2 - 1(n+1)^2 Combining the non-bn terms over the common denominator 2n^2(n+1)^2, bn+1 = 12bn + (n+1)^2 + (n^2-2n-1) - 2n^22n^2(n+1)^2 bn+1 = 12bn Now, b1 = a1 - 11^2 = 1 - 1 = 0 Hence bn=0 for all n 1, so an = 1n^2 Therefore, n=1^ (an - 2n^2) = n=1^ ( 1n^2 - 2n^2) = - n=1^ 1n^2 = - ^26 The solution concludes that the series sum is - ^26, while the answer key is 3. Since this is a numerical value answer and the extracted solution does not support an integer answer, the final answer is marked AMBIGUOUS.

Let a1=1 and for n 1, an+1 = 12an + n^2-2n-12n^2(n+1)^2. Then n=1… — Solution | JEE Mathematics

NVAMediumJEE 2026Sum of Series

Let $a_1=1$ and for $n \ge 1$ , $a_{n+1} = \frac{1}{2}a_n + \frac{n^2-2n-1}{2n^2(n+1)^2}$ . Then $\sum_{n=1}^\infty \left(a_n - \frac{2}{n^2}\right)$ is equal to:

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given: $a_1=1$ and for $n \ge 1$ ,

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

Find: $\sum_{n=1}^\infty \left(a_n - \frac{2}{n^2}\right)$

Define a new sequence

b_n = a_n - \frac{1}{n^2}

Then

a_n = b_n + \frac{1}{n^2}, \qquad a_{n+1} = b_{n+1} + \frac{1}{(n+1)^2}

Substituting into the recurrence,

b_{n+1} + \frac{1}{(n+1)^2} = \frac{1}{2}\left(b_n + \frac{1}{n^2}\right) + \frac{n^2-2n-1}{2n^2(n+1)^2}

So,

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

Combining the non- $b_n$ terms over the common denominator $2n^2(n+1)^2$ ,

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

b_{n+1} = \frac{1}{2}b_n

Now,

b_1 = a_1 - \frac{1}{1^2} = 1 - 1 = 0

Hence $b_n=0$ for all $n \ge 1$ , so

a_n = \frac{1}{n^2}

Therefore,

\sum_{n=1}^\infty \left(a_n - \frac{2}{n^2}\right) = \sum_{n=1}^\infty \left(\frac{1}{n^2} - \frac{2}{n^2}\right) = -\sum_{n=1}^\infty \frac{1}{n^2} = -\frac{\pi^2}{6}

The solution concludes that the series sum is $-\frac{\pi^2}{6}$ , while the answer key is $3$ . Since this is a numerical value answer and the extracted solution does not support an integer answer, the final answer is marked AMBIGUOUS.

Common mistakes

Defining the transformed sequence with the wrong sign. If you take $b_n = a_n + \frac{1}{n^2}$ , the recurrence does not simplify correctly. Instead, choose $b_n = a_n - \frac{1}{n^2}$ so the extra rational terms cancel.
Stopping after obtaining $b_{n+1}=\frac{1}{2}b_n$ and assuming a nonzero geometric progression. Since $b_1=0$ , every term is actually $0$ . Always use the initial condition before writing the final form of $a_n$ .
Using the answer key without checking the recurrence solution. The worked solution gives $a_n=\frac{1}{n^2}$ , which makes the series equal to $-\frac{\pi^2}{6}$ , not an integer. Always verify the final value from the derived expression.

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