Calculate the summations

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- Sep 18th 2012, 01:21 PMWORLDCalculate the summations
Calculate the summations

- Sep 18th 2012, 04:25 PMtopsquarkRe: Calculate the summations
How far have you been able to work these? And are these 4 questions?

-Dan - Sep 19th 2012, 09:36 AMWORLDRe: Calculate the summations
the firs problem

- Sep 19th 2012, 09:37 AMWORLDRe: Calculate the summations
the second problem

- Sep 19th 2012, 09:38 AMWORLDRe: Calculate the summations
the third problem

- Sep 19th 2012, 09:38 AMWORLDRe: Calculate the summations
the last problem

- Sep 19th 2012, 09:43 AMWORLDRe: Calculate the summations
till the farther far

- Sep 19th 2012, 11:52 AMPetrusRe: Calculate the summations
i would also intressted on how to solve this :D

- Sep 19th 2012, 09:49 PMtopsquarkRe: Calculate the summations
- Sep 19th 2012, 10:04 PMWORLDRe: Calculate the summations
It’s mean

- Sep 20th 2012, 11:41 PMWORLDRe: Calculate the summations
could any one help me

- Sep 22nd 2012, 05:12 PMawkwardRe: Calculate the summations
Hello, WORLD!

This is a tough problem for the algebra forum. I'm going to use a little calculus. I hope you will still find my remarks useful.

Start with the Binomial Theorem:

$\displaystyle (1+x)^n = \sum_{i=0}^n \binom{n}{i} x^i$

Integrate both sides from 0 to x (this is the calculus part). The result is

$\displaystyle \frac{1}{n+1}[(1+x)^{n+1}-1] = \sum_{i=0}^n \frac{1}{i+1} \binom{n}{i} x^{i+1}$

Now multiply these two equations together. I'm going to replace i with j in the first equation, so it will be easier to keep everything straight.

(*)....$\displaystyle (1+x)^n \cdot \frac{1}{n+1}[(1+x)^{n+1}-1] = \sum_{j=0}^n \binom{n}{j} x^j \cdot \sum_{i=0}^n \frac{1}{i+1} \binom{n}{i} x^{i+1}$

Now consider the coefficient of $\displaystyle x^{n+1}$ in (*). The left-hand side of (*) is equal to

$\displaystyle \frac{1}{n+1}[(1+x)^{2n+1} - (1+x)^{n}] = \frac{1}{n+1} \left[ \sum_{i=0}^{2n+1} \binom{2n+1}{i} x^i - \sum_{i=0}^n \binom{n}{i}x^i \right] $

so the coefficient of $\displaystyle x^{n+1}$ is

$\displaystyle \frac{1}{n+1} \binom{2n+1}{n+1}$

On the right-hand side of (*), the coefficient of $\displaystyle x^{n+1}$ occurs when j = n-i, so the coefficient is

$\displaystyle \sum_{i=0}^n \binom{n}{n-i} \cdot \frac{1}{i+1} \binom{n}{i} = \sum_{i=0}^n \frac{1}{i+1} \binom{n}{i}^2$

since $\displaystyle \binom{n}{i} = \binom{n}{n-i}$.

The coefficient of $\displaystyle x^{n+1}$ must be the same on both sides of the equation, so we have

$\displaystyle \frac{1}{n+1} \binom{2n+1}{n+1} = \sum_{i=0}^n \frac{1}{i+1} \binom{n}{i}^2$

Now set n = 100, and you're done.