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If (1 + x)^(n) = C(0) + C(1)x + C(2) x^...

If ` (1 + x)^(n) = C_(0) + C_(1)x + C_(2) x^(2) + c_(3) x^(3) + …+ C_(n) x^(n)`, show that
`sum_(r=0)^(n) (C_(r) 3^(r+4))/((r+1)(r+2)(r+3)(r+4))` =
` (1)/((n+1)(n+2)(n+3)(n+4))(4^(n+4) -sum_(t=0)^(3) ""^(n+4)C_(t))`.

Text Solution

Verified by Experts

`LHS = sum_(r=0)^(n) (C_(r) 3^(r+4))/((r+1)(r+2)(r+3)(r+4))`
` = sum_(r=0)^(n)((C_(r)*3^(r+4))/((n+1)(n+2)(n+3)(n+4)))/(4!) 4!`
`= sum_(r=0)^(n) (C_(r) 3^(r+4))/(""^(r+4)C_(4) *4!)= sum_(r=0)^(n)(n!)/(!(n-r)!) *(3^(r+4))/(((r+4)!)/(4!r!)*4!)`
`= sum_(r=0)^(n) (n!*3^(r+4))/((n-r)!*(n+4)!)`
`=sum_(r=0)^(n) (n!*3^(r+4))/((n-r)!*(n+4)!)*((n+1)(n+2)(n+3)(n+4))/((n+1)(n+2)(n+3)(n+4))`
`=sum_(r=0)^(n) (n!*3^(r+4))/((n-r)!*(n+4)!(n+1)(n+2)(n+3)(n+4))`
`=(1)/((n+1)(n+2)(n+3)(n+4))[sum_(r=0)^(n) ((n+4)!*3^(*r+4))/((n-r)!*(r+4)!)]`
`=(1)/((n+1)(n+2)(n+3)(n+4))[sum_(r=0)^(n)""^(n+4)C_(r+4) 3^(r+4)]`
`=(1)/((n+1)(n+2)(n+3)(n+4)){sum_(r=0)^(n)""^(n+4)C_(t)3^(t)}`[put r + 4 = t]
`=(1)/((n+1)(n+2)(n+3)(n+4)){sum_(r=0)^(n)""^(n+4)C_(t)3^(t)- sum_(t=0)^(3) ""^(3^(t)n+4)C_(t)3^(t)} `
`=(1)/((n+1)(n+2)(n+3)(n+4)){(1+3)^(n+4) sum_(t=0)^(3) ""^(3^(t)n+4)C_(t)3^(t)}`
`=(1)/((n+1)(n+2)(n+3)(n+4)){4^(n+4)- sum_(t=0)^(3) ""^(n+4)C_(t)3^(t)}`
RHS .
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