Sum it up

Calculus Level 4

Find the sum of the series:

2 2 1.2 × n C 0 + 2 3 2.3 × n C 1 + 2 4 3.4 × n C 2 + 2 5 4.5 × n C 3 + . . . . . n t e r m s \frac { { 2 }^{ 2 } }{ 1.2 } \times ^{ n }{ C }_{ 0 }+\frac { { 2 }^{ 3 } }{ 2.3 } { \times ^{ n }C }_{ 1 }+\frac { { 2 }^{ 4 } }{ 3.4 } { \times ^{ n }C }_{ 2 }+\frac { { 2 }^{ 5 } }{ 4.5 } { \times ^{ n }C }_{ 3 }+.\quad .\quad .\quad .\quad .\quad n\quad terms

Details and Assumptions:

n C r = n ! ( n − r ) ! r ! ^{ n }{ C }_{ r }=\frac { n! }{ (n-r)!r! }

1 n ( 3 n + 1 n + 1 − 2 ) \frac { 1 }{ n } (\frac { { 3 }^{ n+1 } }{ n+1 } -2) 1 n ( 3 n n − 2 ) \frac { 1 }{ n } (\frac { { 3 }^{ n } }{ n } -2) 1 n + 1 ( 3 n + 1 n + 1 − 2 ) \frac { 1 }{ n+1 } (\frac { { 3 }^{ n+1 } }{ n+1 } -2) 1 n + 1 ( 3 n + 2 n + 2 − 2 ) \frac { 1 }{ n+1 } (\frac { { 3 }^{ n+2 } }{ n+2 } -2)

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Yash Choudhary by Yash Choudhary , 22, India

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1 solution

Kartik Sharma
Mar 4, 2015

Just consider the binomial expansion of ( 1 + x ) n {(1+x)}^{n} and integrate it twice!

*And then of course substitute x = 2 x = 2

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