Sequences & SerieseasyFree

AP Sum cn(n-1): Sum of Squares of Terms | JEE

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

Question
If the sum of nn terms of an A.P. is cn(n−1)cn(n-1) with c≠0c\ne 0, then the sum of the squares of these terms is
Ac2n2(n+1)2c^2n^2(n+1)^2
B23c2n(n−1)(2n−1)\frac23 c^2 n(n-1)(2n-1)correct
C2c23n(n+1)(2n+1)\frac{2c^2}{3}n(n+1)(2n+1)
D2c2(n−1)n(n+1)(2n−3)3\frac{2c^2(n-1)n(n+1)(2n-3)}{3}
Solution
Step 1: The nnth term is
Tn=Sn−Sn−1=cn(n−1)−c(n−1)(n−2)=2c(n−1).T_n=S_n-S_{n-1}=cn(n-1)-c(n-1)(n-2)=2c(n-1).
 Step 2: Sum of squares:
∑k=1n[2c(k−1)]2=4c2∑k=1n(k−1)2=4c2⋅(n−1)n(2n−1)6.\sum_{k=1}^{n}\big[2c(k-1)\big]^2=4c^2\sum_{k=1}^{n}(k-1)^2=4c^2\cdot\frac{(n-1)n(2n-1)}{6}.
 Step 3: =23c2n(n−1)(2n−1).=\dfrac23 c^2 n(n-1)(2n-1). Correct answer: (2)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Sequences & Series · easy
a, b, c, d are in increasing G.P. If the A.M. between a and b is 6 and the A.M. between c…
Sequences & Series · medium
Let a₁, a₂, …, a(4001) be in A.P. such that (1)/(a₁ a₂)+(1)/(a₂ a₃)+⋯+ 1 a(4000) a(4001)…
Sequences & Series · medium
If (p+q· 5^x)/(p-q· 5^x)=(q+r· 5^x)/(q-r· 5^x)=(r+s· 5^x)/(r-s· 5^x) , then p, q, r, s ar…
Sequences & Series · easy
Let a₁, a₂, … and b₁, b₂, … be arithmetic progressions with a₁=25 , b₁=75 and a(100)+b(10…

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.

Sign up freeExplore doMath