JEE Main · Numerical Answer Type · Sequences & Series
Numerical Answer · Mathematics · Sequences & Series
Q. Let $a_1 = 1$ and for $n \geqslant 1$,
$$a_{n+1} = \frac{1}{2}a_n + \frac{n^2 - 2n - 1}{n^2(n+1)^2}$$
Then $\left|\displaystyle\sum_{n=1}^{\infty}\!\left(a_n - \dfrac{2}{n^2}\right)\right|$ is equal to _____ .
✅ Correct Numerical Answer
2
Step-by-Step Solution
Substitution
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New Recurrence
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GP
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Infinite Sum
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Answer
1
Let $b_n = a_n - \dfrac{2}{n^2}$ (Key Substitution)
Then $a_n = b_n + \dfrac{2}{n^2}$.
Substitute into the recurrence $a_{n+1} = \dfrac{1}{2}a_n + \dfrac{n^2-2n-1}{n^2(n+1)^2}$:
Substitute into the recurrence $a_{n+1} = \dfrac{1}{2}a_n + \dfrac{n^2-2n-1}{n^2(n+1)^2}$:
$b_{n+1} + \dfrac{2}{(n+1)^2} = \dfrac{1}{2}\!\left(b_n + \dfrac{2}{n^2}\right) + \dfrac{n^2-2n-1}{n^2(n+1)^2}$
2
Simplify the RHS
$b_{n+1} + \dfrac{2}{(n+1)^2} = \dfrac{b_n}{2} + \dfrac{1}{n^2} + \dfrac{n^2-2n-1}{n^2(n+1)^2}$
Combine the last two terms on RHS:
$\dfrac{1}{n^2} + \dfrac{n^2-2n-1}{n^2(n+1)^2}$
$= \dfrac{(n+1)^2 + n^2-2n-1}{n^2(n+1)^2}$
$= \dfrac{n^2+2n+1+n^2-2n-1}{n^2(n+1)^2}$
$= \dfrac{2n^2}{n^2(n+1)^2} = \dfrac{2}{(n+1)^2}$
$= \dfrac{(n+1)^2 + n^2-2n-1}{n^2(n+1)^2}$
$= \dfrac{n^2+2n+1+n^2-2n-1}{n^2(n+1)^2}$
$= \dfrac{2n^2}{n^2(n+1)^2} = \dfrac{2}{(n+1)^2}$
3
The $\dfrac{2}{(n+1)^2}$ cancels from both sides
$b_{n+1} + \dfrac{2}{(n+1)^2} = \dfrac{b_n}{2} + \dfrac{2}{(n+1)^2}$
$\Rightarrow b_{n+1} = \dfrac{1}{2}\,b_n$
$\Rightarrow b_{n+1} = \dfrac{1}{2}\,b_n$
✅ $\{b_n\}$ is a Geometric Progression with common ratio $r = \dfrac{1}{2}$
4
Find $b_1$
$b_1 = a_1 - \dfrac{2}{1^2} = 1 - 2 = -1$
So the GP is: $b_n = -1 \cdot \left(\dfrac{1}{2}\right)^{n-1} = -\dfrac{1}{2^{n-1}}$
5
Sum the infinite GP $\displaystyle\sum_{n=1}^{\infty} b_n$
First term $b_1 = -1$, ratio $r = \dfrac{1}{2}$, $|r| < 1$ ✓
$\displaystyle\sum_{n=1}^{\infty} b_n = \dfrac{b_1}{1-r} = \dfrac{-1}{1 - \frac{1}{2}} = \dfrac{-1}{\frac{1}{2}} = -2$
6
Apply absolute value
$\left|\displaystyle\sum_{n=1}^{\infty}\!\left(a_n - \dfrac{2}{n^2}\right)\right| = \left|\displaystyle\sum_{n=1}^{\infty} b_n\right| = |-2|$
$= \mathbf{2}$
Related Theory
📌 Particular Solution Method for Recurrences
A recurrence of the form $a_{n+1} = c\cdot a_n + f(n)$ is a first-order linear recurrence. The substitution $b_n = a_n - p_n$, where $p_n$ is a particular solution, converts it to a homogeneous recurrence $b_{n+1} = c\cdot b_n$.
In this problem, $p_n = \dfrac{2}{n^2}$ is the particular solution. The verification: $$\frac{2}{(n+1)^2} = \frac{1}{2}\cdot\frac{2}{n^2} + \frac{n^2-2n-1}{n^2(n+1)^2}$$ is confirmed in Step 2, making $b_n = a_n - 2/n^2$ a pure GP.
In this problem, $p_n = \dfrac{2}{n^2}$ is the particular solution. The verification: $$\frac{2}{(n+1)^2} = \frac{1}{2}\cdot\frac{2}{n^2} + \frac{n^2-2n-1}{n^2(n+1)^2}$$ is confirmed in Step 2, making $b_n = a_n - 2/n^2$ a pure GP.
📌 Infinite GP Sum Formula
For an infinite GP with first term $a$ and common ratio $r$ where $|r| < 1$:
$S_\infty = \dfrac{a}{1-r}$
Here: $a = b_1 = -1$, $r = 1/2$:
$$S_\infty = \frac{-1}{1 - \frac{1}{2}} = \frac{-1}{\frac{1}{2}} = -2$$
The condition $|r| = 1/2 < 1$ ensures convergence.
📌 Algebraic Simplification — Combining Fractions
The critical step was combining:
$$\frac{1}{n^2} + \frac{n^2-2n-1}{n^2(n+1)^2} = \frac{(n+1)^2 + n^2-2n-1}{n^2(n+1)^2}$$
Numerator: $(n^2+2n+1) + (n^2-2n-1) = 2n^2$
Result: $\dfrac{2n^2}{n^2(n+1)^2} = \dfrac{2}{(n+1)^2}$ ✓
This elegant cancellation is the heart of the problem. Recognising that the non-homogeneous terms cancel is the key insight.
Result: $\dfrac{2n^2}{n^2(n+1)^2} = \dfrac{2}{(n+1)^2}$ ✓
This elegant cancellation is the heart of the problem. Recognising that the non-homogeneous terms cancel is the key insight.
📌 Common Mistakes to Avoid
❌ Mistake 1: Not guessing the particular solution $p_n = 2/n^2$ — try $p_n = A/n^2$ and solve for $A$.
❌ Mistake 2: Forgetting the absolute value — the sum is $-2$ but the answer asks for $|-2| = 2$.
❌ Mistake 3: Using the finite GP sum formula instead of infinite GP formula. Since $n \to \infty$, use $S = a/(1-r)$.
❌ Mistake 4: Not verifying $|r| < 1$ before applying the infinite GP formula. Here $r = 1/2$, so it converges.
❌ Mistake 2: Forgetting the absolute value — the sum is $-2$ but the answer asks for $|-2| = 2$.
❌ Mistake 3: Using the finite GP sum formula instead of infinite GP formula. Since $n \to \infty$, use $S = a/(1-r)$.
❌ Mistake 4: Not verifying $|r| < 1$ before applying the infinite GP formula. Here $r = 1/2$, so it converges.
📌 Key Formulas
Infinite GP: $S = \dfrac{a}{1-r}$, $|r|<1$
GP $n$-th term: $b_n = b_1 \cdot r^{n-1}$
$(n+1)^2 = n^2+2n+1$
Recurrence: $a_{n+1}=ca_n+f(n)$
Sub: $b_n = a_n - p_n$
📌 JEE Relevance
This problem tests recurrence relations combined with infinite series — a pattern increasingly seen in JEE Main NAT questions. The strategy of guessing a particular solution to simplify a recurrence is a powerful technique. 70% of users got this wrong, making it a high-discriminating question. Mastering this substitution method significantly boosts JEE rank.
Frequently Asked Questions
1. What substitution is used to simplify this recurrence?
$b_n = a_n - \dfrac{2}{n^2}$. This converts the non-homogeneous recurrence into $b_{n+1} = \dfrac{1}{2}b_n$.
2. Why is $2/n^2$ chosen as the particular solution?
Because substituting $p_n = 2/n^2$ into the recurrence makes both sides equal, confirming it satisfies the recurrence exactly.
3. What type of sequence is $\{b_n\}$?
A Geometric Progression (GP) with first term $b_1 = -1$ and common ratio $r = 1/2$.
4. What is $b_1$?
$b_1 = a_1 - 2/1^2 = 1 - 2 = -1$.
5. What is the infinite sum $\sum b_n$?
$\sum_{n=1}^{\infty} b_n = \dfrac{-1}{1 - 1/2} = -2$.
6. Why is the absolute value taken?
The problem asks for $\left|\sum(a_n - 2/n^2)\right|$. Since the sum equals $-2$, the absolute value gives $|-2| = 2$.
7. Does the infinite series converge?
Yes. The GP has ratio $r = 1/2$ with $|r| < 1$, so the infinite sum converges.
8. What is the general term $b_n$?
$b_n = -1 \cdot (1/2)^{n-1} = -(1/2)^{n-1}$.
9. What is the final numerical answer?
The answer is $\mathbf{2}$.
10. What percentage of users got this wrong?
70% of users answered incorrectly, making it a high-difficulty JEE Main question.
11. What is the infinite GP formula?
$S_\infty = \dfrac{a}{1-r}$ for $|r| < 1$, where $a$ is the first term and $r$ is the common ratio.
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Numerical Answer Type
✅ Correct Answer: 1
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(A) $\dfrac{4}{e^{\pi/4}} - \dfrac{\pi}{2} - 1$
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(C) $\dfrac{4}{e^{\pi/4}} + \dfrac{\pi}{2} - 1$
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(C) 3248
(D) 3244
✅ Correct Answer: (B) 3252