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8. Sequences and Series
normal
If the sum of first $n$ terms of an $A.P.$ is $cn(n -1)$ , where $c \neq 0$ , then sum of the squares of these terms is
A
$c^2n^2(n+1)^2$
B
$\frac{2}{3}c^2n(n-1)(2n-1)$
C
$\frac{2}{3}c^2n(n+1)(2n+1)$
D
$\frac{c^2 n^2}{3}(n+1)^2$
Solution
$\mathrm{S}_{\mathrm{n}}=\mathrm{cn}(\mathrm{n}-1)$
$S_{n-1}=c(n-1)(n-2)$
$t_{n}=S_{n}-S_{n}=c(n-1)(n-(n-2))=2 c(n-1)$
So, for the new series
$\mathrm{T}_{\mathrm{n}}=4 \mathrm{c}^{2}(\mathrm{n}-1)^{2}=4 \mathrm{c}^{2}\left[\mathrm{n}^{2}-2 \mathrm{n}+1\right]$
So,${S_n}$ for the new series will give
${S_n} = 4{c^2}\left[ {\sum {{n^2} – 2\sum {n + n} } } \right] = \frac{2}{3}{c^2}n(n – 1)(2n – 1)$
Standard 11
Mathematics
