# [SOLVED] Using Binomial coefficients in expansion of (1 + x)^n, prove that...?

• Apr 30th 2009, 08:08 PM
fardeen_gen
[SOLVED] Using Binomial coefficients in expansion of (1 + x)^n, prove that...?
If $\displaystyle {n\choose 0}, {n\choose 1}, {n\choose 2}, ... , {n\choose n}$ denote the binomial coefficients in the expansion of $\displaystyle (1 + x)^n$, prove that $\displaystyle \sum_{r = 0}^{n} (-1)^r {n\choose r}\frac{1 + r\log_{e} 10}{(1 + \log_{e} 10^n)^r} = 0$.
• Jun 3rd 2009, 05:14 PM
the_doc
The way I did it:

Spoiler:

Considering the sum

$\displaystyle \sum_{r = 0}^{n} (-1)^r {n\choose r} (1 + r\ln 10) x^r$

which is equal to the given sum when

$\displaystyle x = \frac{1}{1+ n \ln 10}$

we have

$\displaystyle \sum_{r = 0}^{n} (-1)^r {n\choose r} (1 + r\ln 10) x^r = \sum_{r = 0}^{n} (-1)^r {n\choose r} x^r + \ln 10 \, \sum_{r = 0}^{n} (-1)^r {n\choose r} r x^r$,

$\displaystyle = (1-x)^n + \ln 10 \sum_{r = 0}^{n} (-1)^r {n\choose r} r x^r$,

$\displaystyle = (1-x)^n + \ln 10 \, \, x \frac{d}{dx} \left( \sum_{r = 0}^{n} (-1)^r {n\choose r} x^r \right)$,

$\displaystyle = (1-x)^n + \ln 10 \, \, x \frac{d}{dx} \left( (1-x)^n \right)$,

$\displaystyle = (1-x)^n - n x (1-x)^{n-1} \ln 10$,

$\displaystyle = (1-x)^{n-1} \left( 1-x - n x \ln 10 \right)$,

$\displaystyle = (1-x)^{n-1} \left( 1- (1+ n \ln 10)x \right)$.

Hence

$\displaystyle \sum_{r = 0}^{n} (-1)^r {n\choose r} \frac{1 + r\ln 10}{(1+n \ln 10)^r} = \left(1-\frac{1}{1+ n \ln 10} \right)^{n-1} \left( 1- (1+ n \ln 10) \cdot \frac{1}{1+ n \ln 10} \right)$,

$\displaystyle = 0$.