A) \[\frac{1}{2}(n-1)n({{n}^{2}}+3n+4)\]
B) \[\frac{1}{4}(n-1)n({{n}^{2}}+3n+4)\]
C) \[\frac{1}{2}(n+1)n({{n}^{2}}+3n+4)\]
D) \[\frac{1}{4}(n+1)n({{n}^{2}}+3n+4)\]
Correct Answer: B
Solution :
\[{{r}^{th}}\]term of the given series = \[r[(r+1)-\omega ][(r+1)-{{\omega }^{2}}]\] = \[r[{{(r+1)}^{2}}-(\omega +{{\omega }^{2}})(r+1)+{{\omega }^{3}}]\] = \[r[{{(r+1)}^{2}}-(-1)(r+1)+1]\] = \[r[({{r}^{2}}+3r+3]={{r}^{3}}+3{{r}^{2}}+3r\] Thus sum of the given series \[=\sum\limits_{r=1}^{(n-1)}{({{r}^{3}}+3{{r}^{2}}+3r)}\] \[=\frac{1}{4}{{(n-1)}^{2}}{{n}^{2}}+3.\frac{1}{6}(n-1)(n)(2n-1)+3.\frac{1}{2}(n-1)n\] \[=\frac{1}{4}(n-1)n({{n}^{2}}+3n+4)\]
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