Sequence And Series Question 85
Question: If the sum of n terms of an A.P. is cn (n - 1), where $ c\ne 0 $ then the sum of the squares of these term is
Options:
A) $ c^{n}n{{(n+1)}^{2}} $
B) $ \frac{2}{3}c^{2}n(n-1)(2n-1) $
C) $ \frac{2c^{2}}{3}n(n+1)(2n+1) $
D) none of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] If $ t_{r} $ be the rth term of the A.P., then $ t_{r}=S_{r}-{S_{r-1}} $ $ =cr(r-1)-c(r-1)(r-2) $ $ =c(r-1)(r-r+2)=2c(r-1) $ We have, $ t_1^{2}+t_2^{2}+…+t_n^{2}=4c^{2}(0^{2}+1^{2}+2^{2}…+{{(n-1)}^{2}}) $ $ =4c^{2}\frac{(n-1)n(2n-1)}{6} $ $ =\frac{2}{3}c^{2}n(n-1)(2n-1) $
BETA
