A) \[\frac{n\,(n+1)(9{{n}^{2}}+23n+13)}{6}\]
B) \[\frac{n\,(n-1)(9{{n}^{2}}+23n+12)}{6}\]
C) \[\frac{(n+1)(9{{n}^{2}}+23n+13)}{6}\]
D) \[\frac{n\,(9{{n}^{2}}+23n+13)}{6}\]
Correct Answer: A
Solution :
Given series is, \[1\ .\ 3\ .\ 5+2\ .\ 5\ .\ 8+3\ .\ 7\ .\ 11+.......+n(2n+1)(3n+2)\ \] So, \[{{T}_{n}}=n(2n+1)(3n+2)=n\,[6{{n}^{2}}+4n+3n+2]\] \[{{T}_{n}}=6{{n}^{3}}+7{{n}^{2}}+2n\] Now, sum \[=6\Sigma {{n}^{3}}+7\Sigma {{n}^{2}}+2\Sigma n\] \[=6{{\left[ \frac{1}{2}n(n+1) \right]}^{2}}+7\left[ \frac{1}{6}n(n+1)(2n+1) \right]+2\left[ \frac{1}{2}n(n+1) \right]\] \[=\frac{1}{6}n(n+1)[9n\,(n+1)+7(2n+1)+6]\] \[=\frac{1}{6}n\,(n+1)[9{{n}^{2}}+9n+14n+7+6]\] \[=\frac{n\,(n+1)(9{{n}^{2}}+23n+13)}{6}\].
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