Question

The sum of first 9 terms of the series (1^3/1)+(1^3+2^3)/(1+3)+(1^3+2^3+3^3/1+3+5)+...... is

Answer 1

Please find below the solution to the asked query:

Consider the following series.   S=131+13+231+3+13+23+331+3+5+⋯⋯Here the nth term is,   an=13+23+33+⋯⋯+n31+3+5+⋯⋯+2n-1The sum of cube of first n natural numbers is,  13+23+33+⋯⋯+n3=nn+122                                      =n2 n+124Now find the sum of first n odd numbers.Note that,    1=12  1+3=4 =22  1+3+5=9=32  1+3+5+7=16 =42    ⋮    ⋮1+3+5+⋯⋯+2n-1=n2So the nth term of the series becomes,   an=13+23+33+⋯⋯+n31+3+5+⋯⋯+2n-1     =n2n+124n2     =n+124     =14n2+2n+1So the required sum is,     S=∑k=1n ak    =∑k=1n14k2+2k+1    =14∑k=1nk2+2k+1    =14∑k=1nk2+2∑k=1nk+∑k=1n1    =14nn+12n+16+2nn+12+n    =14n n+12n+16+n+1+1    =14n 2n2+3n+1+6n+26    =124n 2n2+9n+13

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