11th Class Mathematics Other Series Question Bank Critical Thinking

  • question_answer
    For any odd integer \[n\ge 1\],\[{{n}^{3}}-{{(n-1)}^{3}}+...........+{{(-1)}^{n-1}}{{1}^{3}}=\] [IIT 1996]

    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) \[\frac{1}{4}{{(n+1)}^{2}}(2n-1)\]

    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 = \[{{n}^{3}}-{{(n-1)}^{3}}+{{(n-2)}^{3}}-{{(n-3)}^{3}}\]+\[........+{{(-1)}^{n-1}}{{1}^{3}}\] \[={{n}^{3}}+{{(n-1)}^{3}}\]\[+{{(n-2)}^{3}}+.......+{{1}^{3}}\]                                    \[-2[{{(n-1)}^{3}}+{{(n-3)}^{3}}+.......+{{2}^{3}}]\] \[={{n}^{3}}+{{(n-1)}^{3}}+{{(n-2)}^{3}}+.......+{{1}^{3}}\]                    \[-2\times {{2}^{3}}\left[ {{\left( \frac{n-1}{2} \right)}^{3}}+{{\left( \frac{n-3}{2} \right)}^{3}}+.......+{{1}^{3}} \right]\]                                    [\[\because \ n-1,\ n-3\]are even integers] \[={{\left[ \frac{n\,(n+1)}{2} \right]}^{2}}-16{{\left[ \frac{1}{2}\left( \frac{n-1}{2} \right)\,\left( \frac{n-1}{2}+1 \right) \right]}^{2}}\] \[=\frac{1}{4}{{n}^{2}}{{(n+1)}^{2}}-16\frac{{{(n-1)}^{2}}{{(n+1)}^{2}}}{16\times 4}\] \[=\frac{1}{4}{{(n+1)}^{2}}[{{n}^{2}}-{{(n-1)}^{2}}]=\frac{1}{4}{{(n+1)}^{2}}(2n-1)\].


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