1.

`sum_(r=1)^(n)(3^(r)-2^(r))=.............`

Answer» Correct Answer - A::B::C
`sum_(r=1)^(n)(3^(r)-2^(r))=sum_(r=1)^(n)3^(r)-sum_(r=1)^(n)2r`
`=(3+3^(2)+3^(3)+.........+3^(n))=(3(3^(n)-1))/(3-1)=(2(2^(n)-1))/(2-1)`
`=(3^(n+1))/(2)-(3)/(2)-2^(n+1)+2(2+2^(2)+2^(3)+..........+2^(n))`
`=(3^(n+1))/(2)-2^(n+1)+(1)/(2)=(3^(n+1)-2^(n+2)+1)/(2)`


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