If the sum of first terms of an AP is , then the sum of squares of these terms is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
The nth term is aₙ = S(n) - S(n-1) = cn² - c(n-1)² = c(2n-1).
Sum of squares = c² Σ(2k-1)² from k=1 to n = c² Σ(4k²-4k+1).
= c²[4·n(n+1)(2n+1)/6 - 4·n(n+1)/2 + n]
= c²·n[4(n+1)(2n+1)/6 - 2(n+1) + 1]
= c²·n[(4(2n²+3n+1)/6) - 2n - 2 + 1]
= c²·n[(8n²+12n+4)/6 - 2n - 1]
= c²·n[(8n²+12n+4 - 12n - 6)/6]
= c²·n[(8n²-2)/6]
= c²·n·2(4n²-1)/6
= n(4n²-1)c²/3.
Sum of squares = c² Σ(2k-1)² from k=1 to n = c² Σ(4k²-4k+1).
= c²[4·n(n+1)(2n+1)/6 - 4·n(n+1)/2 + n]
= c²·n[4(n+1)(2n+1)/6 - 2(n+1) + 1]
= c²·n[(4(2n²+3n+1)/6) - 2n - 2 + 1]
= c²·n[(8n²+12n+4)/6 - 2n - 1]
= c²·n[(8n²+12n+4 - 12n - 6)/6]
= c²·n[(8n²-2)/6]
= c²·n·2(4n²-1)/6
= n(4n²-1)c²/3.
[AIEEE 2009]