A) 188090
B) 189080
C) 199080
D) None of these
Correct Answer: A
Solution :
Here \[{{T}_{n}}\] of the A.P. \[1,\ 2,\ 3,\ ..........=n\] and \[{{T}_{n}}\] of the A.P. \[3,\ 5,\ 7,........=2n+1\] \[\therefore \]\[{{T}_{n}}\] of given series \[=n{{(2n+1)}^{2}}=4{{n}^{3}}+4{{n}^{2}}+n\] Hence \[S=\sum\limits_{1}^{20}{{{T}_{n}}}=4\sum\limits_{1}^{20}{{{n}^{3}}}+4\sum\limits_{1}^{20}{{{n}^{2}}}+\sum\limits_{1}^{20}{n}\] \[=4\cdot \frac{1}{4}{{20}^{2}}\cdot {{21}^{2}}+4\cdot \frac{1}{6}20\cdot 21\cdot 41+\frac{1}{2}20\cdot 21=188090\]
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