Theory of EquationsmediumFree

Telescoping Recurrence a(n) - a(n-1) = n^2 with a(k) = 205 | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question

Let

a_1, a_2, \ldots, a_n

be a sequence such that

a_1 = 2

and

a_n - a_{n-1} = n^{2} \quad \forall\ n \ge 2 .

a_k = 205

, then

k

is equal to

5

8

correct

11

15

Solution

Step 1: Write the recurrence out for every index from

2

up to

k

a_2 - a_1 = 2^{2}, \quad a_3 - a_2 = 3^{2}, \quad \ldots, \quad a_k - a_{k-1} = k^{2} .

Step 2: Add all these equations. Everything on the left cancels in pairs except the ends (a telescoping sum):

a_k - a_1 = 2^{2} + 3^{2} + \cdots + k^{2} .

Step 3: Complete the square-sum by putting back the missing

1^{2}

a_k - a_1 = \left(1^{2}+2^{2}+\cdots+k^{2}\right) - 1 .

Step 4: Substitute

a_1 = 2

and use the standard formula

\sum_{i=1}^{k} i^{2} = \dfrac{k(k+1)(2k+1)}{6}

a_k = 2 + \frac{k(k+1)(2k+1)}{6} - 1 = 1 + \frac{k(k+1)(2k+1)}{6} .

Step 5: Set

a_k = 205

1 + \frac{k(k+1)(2k+1)}{6} = 205 \implies \frac{k(k+1)(2k+1)}{6} = 204 \implies k(k+1)(2k+1) = 1224 .

Step 6: Solve by trying the given options. Since

k(k+1)(2k+1)

grows fast, test

k = 8

8 \times 9 \times 17 = 1224 .

Step 7: Check by direct substitution:

a_8 = 1 + \dfrac{1224}{6} = 1 + 204 = 205

. Correct. Answer: (2).

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.