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In the expansion of (1+x)^n, the successive coefficients are a₀, a₁, a₂, a₃,......,aₙ; show that a₀+ 2a₁+ 3a₂+......+(n+1)aₙ = 2^n + n*2^(n-1)

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8 months ago

2074 Points
```							let (1+x)^n= a0 + a1*x^1 + a2*x^2 + ….... + an*x^n........(1)multiply by x on both sidesx(1+x)^n= a0*x^1 + a1*x^2 + a2*x^3 + ….... + an*x^(n+1)differentiate both sides wrt x(1+x)^n + nx(1+x)^(n-1) = a0 + 2a1*x^1 + 3a2*x^2 + …..... + (n+1)an*x^nput x=1, we get2^n + n*2^(n-1)= a₀+ 2a₁+ 3a₂+......+(n+1)aₙhence proved.KINDLY APPROVE :))
```
8 months ago
Vikas TU
14146 Points
```							Given, (1+x)^n.(1+y)^n.(1+z)^n =[(1+x)(1+x)...(1+x)]n factors=(1+y)(1+y)+....(1+y)]n factors=[(1+z)(1+z)...(1+)z]n factorsThere are 3n factors in (1+x)^n.(1+y)^n.(1+z)^n  for term of degree r. We choose r brackets out of 3n brackets and then multiply second terms in each bracket.There are 3nCr such terms each having coefficient 1. Hence, the sum of the coefficients is  3nCr.
```
8 months ago
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