# Thread: Help with Newton's Binom

1. ## Help with Newton's Binom

Hello, I would be happy to help with the solution,
Simplify the amount: Reached the dependent expression in M, which contains no sums.
thanks 2. ## Re: Help with Newton's Binom

Have you done anything? You should try writing down a few terms explicitly.

3. ## Re: Help with Newton's Binom

I started really a little but I can not move forward and probably it's not true, I would love to get an explanation from the beginning ... 4. ## Re: Help with Newton's Binom

For example, with m= 1 that is 1+ 1/2= 3/2. With m= 2 that is 1+ 1+ 1/3= 7/3. With m= 3 that is 1+ 3/2+ 1+ 1/4= 15/4.

But what I would do is note that $\displaystyle (x+ 1)^m= \sum_{n= 0}^m\begin{pmatrix}m \\ n\end{pmatrix}x^n$ and integrating both sides $\displaystyle \frac{1}{m+ 1}(x+ 1)^{m+1}= \sum_{n= 0}^m \frac{1}{n+1}\begin{pmatrix} m\\ n\end{pmatrix}x^{n+1}$. Now let x= 1.

5. ## Re: Help with Newton's Binom Originally Posted by HallsofIvy For example, with m= 1 that is 1+ 1/2= 3/2. With m= 2 that is 1+ 1+ 1/3= 7/3. With m= 3 that is 1+ 3/2+ 1+ 1/4= 15/4.

But what I would do is note that $\displaystyle (x+ 1)^m= \sum_{n= 0}^m\begin{pmatrix}m \\ n\end{pmatrix}x^n$ and integrating both sides $\displaystyle \frac{1}{m+ 1}(x+ 1)^{m+1}= \sum_{n= 0}^m \frac{1}{n+1}\begin{pmatrix} m\\ n\end{pmatrix}x^{n+1}$. Now let x= 1.
there is a constant of integration so that

$\displaystyle \frac {1} {m + 1} (x + 1)^{m + 1} - \frac {1} {m + 1} = \sum _ {n = 0}^m\frac {1} {n + 1}\binom {m} {n} x^{n + 1}$

6. ## Re: Help with Newton's Binom

\begin{align*}\frac1{n+1}{m \choose n} &= \frac1{n+1} \cdot \frac{m!}{n!(m-n)!} = \frac{m!}{(n+1)!(m-n)!} \\ &= \frac1{m+1}\cdot\frac{(m+1)!}{(n+1)!\big((m+1)-(n+1)\big)!} \\ &= \frac1{m+1}{m+1 \choose n+1}\end{align*}

So \begin{align*}\sum_{n=0}^{m} \frac1{n+1} {m \choose n} &= \sum_{n=0}^{m} \frac1{m+1}{m+1 \choose n+1} = \frac1{m+1} \sum_{n=1}^{m+1} {m+1 \choose n} \\ &= \frac1{m+1}\left[ \left( \sum_{n=0}^{m+1} {m+1 \choose n} \right) - {m+1 \choose 0} \right] \\ &= \frac1{m+1}\left[ \left( \sum_{n=0}^{m+1} {m+1 \choose n} \right) - 1 \right]\end{align*}

Now the summation is simply the sum of all elements in the $(m+1)^\text{th}$ row of Pascal's Triangle (the one with the second element equal to $(m+1)$) and $$\sum_{n=0}^{m+1} {m+1 \choose n} = (1+1)^{m+1} = 2^{m+1}$$

Thus $$\sum_{n=0}^{m} \frac1{n+1} {m \choose n} = \frac1{m+1}\left[ 2^{m+1} - 1 \right]$$

7. ## Re: Help with Newton's Binom

thanks you all 