Sequences & SeriesmediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Series with Σa_k = 6n³: Value of the Squared-Difference Sum = 91 | JEE 2026

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

Question
If ∑k=1nak=6n3\displaystyle\sum_{k=1}^{n}a_k=6n^3, then ∑k=16(ak+1−ak36)2\displaystyle\sum_{k=1}^{6}\left(\dfrac{a_{k+1}-a_k}{36}\right)^2 is equal to
Solution
Answer: 91 (± 0.01)
Step 1: Find ana_n from the partial sums:
an=∑k=1nak−∑k=1n−1ak=6n3−6(n−1)3.a_n=\sum_{k=1}^{n}a_k-\sum_{k=1}^{n-1}a_k=6n^3-6(n-1)^3.
 Expand: 6n3−6(n3−3n2+3n−1)=6(3n2−3n+1)6n^3-6(n^3-3n^2+3n-1)=6(3n^2-3n+1), so
an=18n2−18n+6.a_n=18n^2-18n+6.
 Step 2: Compute the difference:
ak+1−ak=18[(k+1)2−k2]−18[(k+1)−k]=18(2k+1)−18=36k.a_{k+1}-a_k=18\big[(k+1)^2-k^2\big]-18\big[(k+1)-k\big]=18(2k+1)-18=36k.
 Step 3: Then
(ak+1−ak36)2=(36k36)2=k2.\left(\frac{a_{k+1}-a_k}{36}\right)^2=\left(\frac{36k}{36}\right)^2=k^2.
 Step 4: Sum for k=1k=1 to 66:
∑k=16k2=6⋅7⋅136=91.\sum_{k=1}^{6}k^2=\frac{6\cdot7\cdot13}{6}=91.
 Correct answer: 91
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