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 \[=\sum\limits_{r=1}^{(n-1)}{({{r}^{3}}+3{{r}^{2}}+3r)}\] \[=\frac{1}{4}{{(n-1)}^{2}}{{n}^{2}}+3\cdot \frac{1}{6}(n-1)n(2n-1)+3\cdot \frac{1}{2}(n-1)n\]\[=\frac{1}{4}(n-1)n({{n}^{2}}+3n+4)\]
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