Question

Find the sum n terms of the series 11+103+1005+......n?

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{Series\;is\;11+103+1005+10007+..........+n\;terms}$

$\underline{\textbf{To find:}}$

$\textsf{The sum of the given series}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{11+103+1005+10007+..........+n\;terms}$

$\textsf{This can be written as}$

$\mathsf{=(10+1)+(100+3)+(1005)+..........+n\;terms}$

$\textsf{Splitting the series}$

$\mathsf{=(10+100+1000+...........+n\;terms)+(1+3+5+........+n\;terms)}$

$\mathsf{=(10+(10)^2+(10)^3+...........+n\;terms)+(1+3+5+........+n\;terms)}$

$\mathsf{=(Sum\of\;first\;n\;terms\;the\;G.P\,with\,a=10\,and\,r=10)+(Sum\;of\;first\;n\;odd\;natural\;numbers)}$

$\mathsf{=\dfrac{a(r^n-1)}{r-1}+n^2}$

$\mathsf{=\dfrac{10(10^n-1)}{10-1}+n^2}$

$\mathsf{=\dfrac{10}{9}(10^n-1)+n^2}$

$\underline{\textbf{Answer:}}$

$\mathsf{\bf\,11+103+1005+10007+..........+n\;terms=\dfrac{10}{9}(10^n-1)+n^2}$

$\underline{\textbf{Find more:}}$

the third term of G.P. is 3/4 and the sum of the first five terms is 32/33 times the sum of the first ten terms.Find the sum of the first four terms.​

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