A) \[\frac{1}{4}{{(n+1)}^{2}}\,(2n-1)\]
B) \[\frac{1}{4}\,(2n-1){{n}^{2}}\]
C) \[\frac{1}{4}\,{{(n+1)}^{2}}+(2n+1)\]
D) \[\frac{1}{4}\,(2n+1){{n}^{2}}\]
Correct Answer: A
Solution :
[a] Since n is an odd integer, \[{{(-1)}^{n-1}}=1\]and\[n-1,\] \[n-3,\text{ }n-5,\text{ }...\]are even integers. The given series is \[{{n}^{3}}-{{(n-1)}^{3}}+{{(n-2)}^{3}}-{{(n-3)}^{3}}+....+{{(-1)}^{n-1}}{{1}^{3}}\] \[=[{{n}^{3}}+{{(n-1)}^{3}}+{{(n-2)}^{3}}+...+{{1}^{3}}]-2[{{(n-1)}^{3}}\] \[+{{(n-3)}^{3}}+....+{{2}^{3}}]\] \[={{\left[ \frac{n(n+1)}{2} \right]}^{2}}-16{{\left[ \left\{ \frac{1}{2}\left( \frac{n-1}{2} \right)\left( \frac{n-1}{2} \right)+1 \right\} \right]}^{2}}\] \[=\frac{1}{4}{{n}^{2}}{{(n+1)}^{2}}-16\times \frac{{{(n-1)}^{2}}{{(n+1)}^{2}}}{16\times 4}\] \[=\frac{1}{4}{{(n+1)}^{2}}[{{n}^{2}}-{{(n-1)}^{2}}]=\frac{1}{4}{{(n+1)}^{2}}(2n-1)\]
You need to login to perform this action.
You will be redirected in
3 sec
