InterviewSolution
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InterviewSolution
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Find the sum of the series `sum_(r=0)^(n) (-1)^(r ) ""^(n)C_(r ) [(1)/(2^(r )) + (3^(r ))/(2^(2r)) + (7^(r ))/(2^(3 r)) + (15^(r ))/(2^(4r)) …. "upto m terms"]` |
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Answer» `underset (r=0)overset(n)Sigma (-1)^(n) C_(r)[(1)/(2^(r ))+(3^(r ))/(2^(2r))+(7^(r ))/(2^(3r))+(15^(r ))/(2^(4r))+…"upto m terms"]` `=underset(r=0)overset(n)Sigma(-1)^(r ) C_(r)(1/2)^(r )+ underset(r=0)overset(n)Sigma^(r )C_(r )(3/4)^(r )+underset(r=0)overset(n)Sigma(-1)^(r )C_(r )(7/8)^(r )+....` upto m terms `=(1-1/2)^(n)+(1-3/4)^(n)+(1-7/8)^(r) +...` up to m terms `[using underset(r=0)overset(n)Sigma (-1)^(r )C_(r )x^(r )=(1-x)^(n)]` `=(1/2)^(n)+(1/4)^(n)+(1/8)^(n)+...` up to m terms `=(1/2)^(n)[(1=-(1/2)^(m))/(1-1/2)]=(2^(mn)-1)/(2^(mn)(2^(n)-1)`. |
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