binomial q (1 Viewer)

[Master E]

If (1+x)^n = sigma(r=0-->n) (nCr)(x^r) then show by integration that

sigma (r=0-->n) ((-1)^n)(nCr)/(r+1) = 1/(n+1).

Writing it out might be easier to understand.

I integrated both sides, the LHS is easy, the RHS i can get, but i cant work out what to do after that.

Thanks

 

[Trebla]

(1 + x)ⁿ = Σ ⁿC_r x^r
Integrate both sides wrt x
(1 + x)^{n + 1}/(n + 1) = Σ ⁿC_r x^{r + 1}/(r + 1) + c
To find c, let x = 0, which gives c = 1/(n + 1)
(1 + x)^{n + 1}/(n + 1) = Σ ⁿC_r x^{r + 1}/(r + 1) + 1/(n + 1)
Let x = -1
Σ ⁿC_r (-1)^{r + 1}/(r + 1) + 1/(n + 1) = 0
Σ ⁿC_r (-1)(-1)^r/(r + 1) + 1/(n + 1) = 0
- Σ ⁿC_r (-1)^r/(r + 1) + 1/(n + 1) = 0
Σ ⁿC_r (-1)^r/(r + 1) = 1/(n + 1)

I assume you mean (-1)^r rather than (-1)ⁿ...

 

[Master E]

(1 + x)ⁿ = Σ ⁿC_r x^r
Integrate both sides wrt x
(1 + x)^{n + 1}/(n + 1) = Σ ⁿC_r x^{r + 1}/(r + 1) + c
To find c, let x = 0, which gives c = 1/(n + 1)
(1 + x)^{n + 1}/(n + 1) = Σ ⁿC_r x^{r + 1}/(r + 1) + 1/(n + 1)
Let x = -1
Σ ⁿC_r (-1)^{r + 1}/(r + 1) + 1/(n + 1) = 0
Σ ⁿC_r (-1)(-1)^r/(r + 1) + 1/(n + 1) = 0
- Σ ⁿC_r (-1)^r/(r + 1) + 1/(n + 1) = 0
Σ ⁿC_r (-1)^r/(r + 1) = 1/(n + 1)

I assume you mean (-1)^r rather than (-1)ⁿ...

That was the problem! Darn i copied the question out wrong.

Thanks for the help anyhow