Sequence And Series Question 530
Question: For any odd integer $ n\ge 1 $ , $ n^{3}-{{(n-1)}^{3}}+………..+{{(-1)}^{n-1}}1^{3}= $
[IIT 1996]
Options:
A) $ \frac{1}{2}{{(n-1)}^{2}}(2n-1) $
B) $ \frac{1}{4}{{(n-1)}^{2}}(2n-1) $
C) $ \frac{1}{2}{{(n+1)}^{2}}(2n-1) $
D) $ \frac{1}{4}{{(n+1)}^{2}}(2n-1) $
Show Answer
Answer:
Correct Answer: D
Solution:
Since $ n $ is an odd integer $ {{(-1)}^{n-1}}=1 $ and $ n-1 $ , $ n-3,\ n-5 $ etc. are even integers. We have = $ 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}] $ $ =n^{3}+{{(n-1)}^{3}}+{{(n-2)}^{3}}+…….+1^{3} $ $ -2\times 2^{3}[ {{( \frac{n-1}{2} )}^{3}}+{{( \frac{n-3}{2} )}^{3}}+…….+1^{3} ] $ [ $ \because \ n-1,\ n-3 $ are even integers] $ ={{[ \frac{n,(n+1)}{2} ]}^{2}}-16{{[ \frac{1}{2}( \frac{n-1}{2} ),( \frac{n-1}{2}+1 ) ]}^{2}} $ $ =\frac{1}{4}n^{2}{{(n+1)}^{2}}-16\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) $ .
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