Further Binomial Question that I can't do (1 Viewer)

imoO · Apr 27, 2009

Hi guys,

I was wondering if anybody could help me on this question:

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I know there's like three methods, such as differentiating both sides, or finding an appropriate substitution for x, but if someone could guide me through this instead of just giving a solution, that would be great.

Thanks for any help in advanced,

Drongoski · Apr 27, 2009

Here's my attempt

$\100dpi (1+x)^{n+2}\ \ =\ \ (1+x)^{2}(1+x)^{n}\\ \\ Expanding \ both\ sides:\\ \\ \ \ \sum_{r=0}^{n+2}\binom{n+2}{r}\ x^{r}\ \ =\ \ (1+2x+x^{2})\ \sum_{r=0}^{n}\binom{n}{r}\ x^{r}\\ \\ \\ Equating\ coeffs\ of\ x^{r} \ for\ LHS\ and\ RHS:\\ \\ \ \ \binom{n+2}{r}\ \ =\ \ \binom{n}{r}\ +\ 2\ \binom{n}{r-1}\ +\ \binom{n}{r-2}\\ \\ \\ QED$

lychnobity · Apr 27, 2009

n+2Cr is the coefficient of x^r in the expansion of (1 + x)^{n + 2}

(1 + x)^{n + 2} = (1 + x)ⁿ(1 + x)²

=(1 + x)ⁿ(1 + 2x + x²)

For the coefficient of x^r ie:
ⁿC_rx^r + ⁿC_r-1x^r-1(2x) + ⁿC_r-2x^r-2(x²)

If you can read it properly from the above, the powers of x when added always equal to r. Since when finding the coefficient of x^v for (a + bx)ⁿ, the formula is ⁿC_vb^va^n-v

imoO · Apr 29, 2009

Okay,

Thanks guys

Further Binomial Question that I can't do (1 Viewer)

imoO

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Drongoski

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lychnobity

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imoO

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