A) \[{{27}^{2}}\]
B) \[-54\]
C) \[54\]
D) 0
Correct Answer: D
Solution :
[d] General term of the series \[\sum\limits_{n=1}^{10}{\int\limits_{-2n-1}^{-2n}{{{\sin }^{27}}xdx}}\] is \[{{I}_{1}}=\int\limits_{-2n-1}^{-2n}{{{\sin }^{27}}xdx=\int\limits_{2n+1}^{2n}{{{\sin }^{27}}(-x)(-dx)}}\] \[=-\int\limits_{2n}^{2n+1}{{{\sin }^{27}}xdx=-{{I}_{2}}}\] Where \[{{I}_{2}}\] is general term of series \[\sum\limits_{n=1}^{10}{\int\limits_{2n}^{2n+1}{{{\sin }^{27}}xdx}}\] So \[{{I}_{1}}+{{I}_{2}}=0\] for all n
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