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JEE Main 20262 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2026 · 2 April · Morning Shift · Mathematics · Sequences and Series

If ∑_ k=1 ^ n a_k = 6n^3 , then ∑_ k=1 ^ 6 ( a_ k+1 - a_k 36 )^2 is...

If ∑k=1nak=6n3\displaystyle\sum_{k=1}^{n} a_k = 6n^3, then $\displaystyle\sum_{k=1}^{6} \left(\dfrac{a_{k+1} - a_k}{36}\right)^2 is equal to _______.

Correct answer

0

Step-by-step solution

Given ∑k=1nak=6n3\displaystyle\sum_{k=1}^{n} a_k = 6n^3.

Let Sn=6n3S_n = 6n^3. The nn-th term of the sequence is given by an=Sn−Sn−1a_n = S_n - S_{n-1}.

an=6n3−6(n−1)3a_n = 6n^3 - 6(n-1)^3

an=6[n3−(n3−3n2+3n−1)]a_n = 6[n^3 - (n^3 - 3n^2 + 3n - 1)]

an=6(3n2−3n+1)a_n = 6(3n^2 - 3n + 1)

Replacing nn with kk and k+1k+1, we get:

ak=6(3k2−3k+1)a_k = 6(3k^2 - 3k + 1)

ak+1=6(3(k+1)2−3(k+1)+1)=6(3k2+3k+1)a_{k+1} = 6(3(k+1)^2 - 3(k+1) + 1) = 6(3k^2 + 3k + 1)

Now, find the difference ak+1−aka_{k+1} - a_k:

ak+1−ak=6(3k2+3k+1)−6(3k2−3k+1)a_{k+1} - a_k = 6(3k^2 + 3k + 1) - 6(3k^2 - 3k + 1)

ak+1−ak=6(6k)=36ka_{k+1} - a_k = 6(6k) = 36k

Substitute this into the given summation:

∑k=16(ak+1−ak36)2=∑k=16(36k36)2\displaystyle\sum_{k=1}^{6} \left(\dfrac{a_{k+1} - a_k}{36}\right)^2 = \sum_{k=1}^{6} \left(\dfrac{36k}{36}\right)^2

⇒∑k=16k2=12+22+32+42+52+62\Rightarrow \displaystyle\sum_{k=1}^{6} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2

Using the formula for the sum of squares of the first nn natural numbers, ∑k=1nk2=n(n+1)(2n+1)6\displaystyle\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}:

∑k=16k2=6(6+1)(2(6)+1)6=6×7×136=91\displaystyle\sum_{k=1}^{6} k^2 = \dfrac{6(6+1)(2(6)+1)}{6} = \dfrac{6 \times 7 \times 13}{6} = 91

Answer: 9191
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