A) \[\frac{{{n}^{2}}(2n+1)(3n+2)}{6}\]
B) \[\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}\]
C) \[\frac{{{n}^{3}}(n-1)}{3}\]
D) \[\frac{{{(n-3)}^{2}}(n+5)}{9}\]
Correct Answer: B
Solution :
(b): Consider \[{{n}^{4}}-{{(n-1)}^{4}}=4{{n}^{3}}-6{{n}^{2}}+4n-1\] \[\sum{{{n}^{4}}-{{(n-1)}^{4}}=4\sum{{{n}^{3}}-6\sum{{{n}^{2}}+4\sum{n-\sum{1}}}}}\] \[{{n}^{4}}=4s-n(n+1)(2n+1)+2n(n+1)-n\] \[4s={{n}^{4}}+n(n+1)(2n+1)-2n(n+1)+n\] \[4s=\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}\] \[s=\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}\]
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