Question

\[\sum_{r=0}^n (-1)^r \binom{n}{r} \frac{1 + r \ln(10)}{(1 + \ln(10^n))^r}\]

Answer 1

Can't see the question
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Answer 2

\[\sum_{r=0}^n (-1)^r \binom{n}{r} \frac{1 + r \ln(10)}{(1 + \ln(10^n))^r}\]

Answer 3

PK u just reminded me I have some of these to do. ._.

;-;

Answer 4

hey ur alive

Answer 5

just woke up

Answer 6

here's a nice little hint to the previous problem: binomial theorem for any index

Answer 7

which simply put is the maclaurin expansion of \((1+x)^n\)

Answer 8

4:20 am!!!

Answer 9

With
\[(1+x)^n=\sum_{k=0}^n\binom nkx^k~~\implies~~x\frac{d}{dx}(1+x)^n=nx(1+x)^{n-1}=\sum_{k=0}^nk\binom nk x^k\]you can set \(x=-\dfrac{1}{1+n\ln10}\) so that you get
\[\begin{align*}S&=\sum_{r=0}^n (-1)^r \binom{n}{r} \frac{1 + r \ln(10)}{(1 + \ln(10^n))^r}\\[1ex]
&=\left(1-\frac{1}{1+n\ln10}\right)^n-\frac{n\ln10}{1+n\ln10}\left(1-\frac{1}{1+n\ln10}\right)^{n-1}\\[1ex]
&=\left(1-\frac{1}{1+n\ln10}\right)^n\left[1-\frac{n\ln10}{1+n\ln10}\left(\frac{1+n\ln10}{n\ln10}\right)\right]\\[1ex]
&=0
\end{align*}\]

Answer 10

Well done!