A) \[{{c}^{n}}n{{(n+1)}^{2}}\]
B) \[\frac{2}{3}{{c}^{2}}n(n-1)(2n-1)\]
C) \[\frac{2{{c}^{2}}}{3}n(n+1)(2n+1)\]
D) none of these
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, \[{{t}_{1}}^{2}+{{t}_{2}}^{2}+...+{{t}_{n}}^{2}=4{{c}^{2}}({{0}^{2}}+{{1}^{2}}+{{2}^{2}}...+{{(n-1)}^{2}})\] \[=4{{c}^{2}}\frac{(n-1)n(2n-1)}{6}\] \[=\frac{2}{3}{{c}^{2}}n(n-1)(2n-1)\]
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