A) 188090
B) 189080
C) 199080
D) 199089
Correct Answer: A
Solution :
Here\[{{T}_{n}}\]of the \[A.P.\text{ }1,2,3\text{ }...=n\] and\[{{T}_{n}}\]of the \[A.R\text{ }3,5,7...=2n+1\] \[\therefore \] \[{{T}_{n}}\]of given series\[=n(2n+1)\] \[=4{{n}^{3}}+4{{n}^{2}}+n\] hence, \[S=\underset{1}{\overset{2}{\mathop{\Sigma }}}\,{{T}_{n}}=4\underset{1}{\overset{20}{\mathop{\Sigma }}}\,{{n}^{3}}+4\underset{1}{\overset{20}{\mathop{\Sigma }}}\,{{n}^{2}}+\underset{1}{\overset{20}{\mathop{\Sigma }}}\,n\] \[=4\frac{1}{4}{{20}^{2}}{{.21}^{2}}+4\frac{1}{6}20.2141+\frac{1}{2}20.21\]\[=188090\]
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