A) 142
B) 192
C) 71
D) 96
Correct Answer: D
Solution :
\[{{T}_{r}}=\frac{{{1}^{3}}+{{2}^{3}}+.....+{{r}^{3}}}{1+3+5+.......+\left( 2r-1 \right)}=\frac{{{r}^{2}}{{\left( r+1 \right)}^{2}}}{4\left( {{r}^{2}} \right)}=\frac{{{\left( r+1 \right)}^{2}}}{4}\]\[\Rightarrow \]\[sum=\sum\limits_{r=1}^{9}{{{T}_{r}}}=\sum\limits_{r=1}^{n}{\frac{{{\left( r+1 \right)}^{2}}}{4}}\]21 21 \[=\frac{1}{4}\left[ {{2}^{2}}+{{3}^{2}}+.....+{{10}^{2}} \right]=\frac{1}{4}\left[ {{1}^{2}}+{{2}^{2}}+....+{{10}^{2}}-1 \right]=96\]: D /
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