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| 1. |
Find the sum of n-terms of the series 1^2+ +3^2+5^2+........ |
| Answer» Let {tex}{{\\rm{S}}_n} = \\sum {{{\\left( {{\\rm{odd\\ number}}} \\right)}^3}} {/tex}=> {tex}{{\\rm{S}}_n} = \\sum\\limits_{k = 1}^n {\\left( {2k - 1} \\right)} {/tex} [Sn = Sum of n odd natural numbers]=> {tex}{{\\rm{S}}_n} = \\sum\\limits_{k = 1}^n {\\left( {8{k^3} - 12{k^2} + 6x - 1} \\right)} {/tex}=> {tex}{{\\rm{S}}_n} = 8\\left( {\\sum\\limits_{k = 1}^n {{k^3}} } \\right) - 12\\left( {\\sum\\limits_{k = 1}^n {{k^3}} } \\right) + 6\\left( {\\sum\\limits_{k = 1}^n k } \\right) - n{/tex} [1 + 1 + 1 + ...... to n terms = n]=> {tex}{{\\rm{S}}_n} = 8.{1 \\over 4}{n^2}{\\left( {n + 1} \\right)^2} - 12.{1 \\over 6}n\\left( {n + 1} \\right)\\left( {2n + 1} \\right) + 6.{1 \\over 2}n\\left( {n + 1} \\right) - n{/tex}=> {tex}{{\\rm{S}}_n} = 2{n^2}{\\left( {n + 1} \\right)^2} - 2n\\left( {n + 1} \\right)\\left( {2n + 1} \\right) + 3n\\left( {n + 1} \\right) - n{/tex}=> {tex}{{\\rm{S}}_n} = n\\left( {n + 1} \\right)\\left[ {2n\\left( {n + 1} \\right) - 2\\left( {2n + 1} \\right) + 3} \\right] - n{/tex}=> {tex}{{\\rm{S}}_n} = n\\left( {n + 1} \\right)\\left[ {2{n^2} - 2n + 1} \\right] - n{/tex}=> {tex}{{\\rm{S}}_n} = n\\left[ {\\left( {n + 1} \\right)\\left( {2{n^2} - 2n + 1} \\right) - 1} \\right]{/tex}=> {tex}{{\\rm{S}}_n} = n\\left[ {2{n^3} - 2{n^2} + n + 2{n^2} - 2n + 1 - 1} \\right]{/tex}=> {tex}{{\\rm{S}}_n} = n\\left[ {2{n^3} - n} \\right]{/tex}=> {tex}{{\\rm{S}}_n} = {n^2}\\left[ {2{n^2} - 1} \\right]{/tex} | |