Calculate the summations

• September 18th 2012, 01:21 PM
WORLD
Calculate the summations
Calculate the summations
• September 18th 2012, 04:25 PM
topsquark
Re: Calculate the summations
How far have you been able to work these? And are these 4 questions?

-Dan
• September 19th 2012, 09:36 AM
WORLD
Re: Calculate the summations
the firs problem
• September 19th 2012, 09:37 AM
WORLD
Re: Calculate the summations
the second problem
• September 19th 2012, 09:38 AM
WORLD
Re: Calculate the summations
the third problem
• September 19th 2012, 09:38 AM
WORLD
Re: Calculate the summations
the last problem
• September 19th 2012, 09:43 AM
WORLD
Re: Calculate the summations
till the farther far
• September 19th 2012, 11:52 AM
Petrus
Re: Calculate the summations
i would also intressted on how to solve this :D
• September 19th 2012, 09:49 PM
topsquark
Re: Calculate the summations
Quote:

Originally Posted by WORLD
the firs problem

$\Sigma _{k = 0} ^{100} \frac{1}{k + 1} \left ( _k ^{100} \right )$

What does $\left ( _k ^{100} \right )$ mean?

-Dan
• September 19th 2012, 10:04 PM
WORLD
Re: Calculate the summations
It’s mean
• September 20th 2012, 11:41 PM
WORLD
Re: Calculate the summations
could any one help me
• September 22nd 2012, 05:12 PM
awkward
Re: Calculate the summations
Quote:

Originally Posted by WORLD
the firs problem
Find $\sum_{k=0}^{100}\frac{1}{k+1} \binom{100}{k}^2$

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.

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

Integrate both sides from 0 to x (this is the calculus part). The result is
$\frac{1}{n+1}[(1+x)^{n+1}-1] = \sum_{i=0}^n \frac{1}{i+1} \binom{n}{i} x^{i+1}$

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

Now consider the coefficient of $x^{n+1}$ in (*). The left-hand side of (*) is equal to
$\frac{1}{n+1}[(1+x)^{2n+1} - (1+x)^{n}] = \frac{1}{n+1} \left[ \sum_{i=0}^{2n+1} \binom{2n+1}{i} x^i - \sum_{i=0}^n \binom{n}{i}x^i \right]$
so the coefficient of $x^{n+1}$ is
$\frac{1}{n+1} \binom{2n+1}{n+1}$

On the right-hand side of (*), the coefficient of $x^{n+1}$ occurs when j = n-i, so the coefficient is
$\sum_{i=0}^n \binom{n}{n-i} \cdot \frac{1}{i+1} \binom{n}{i} = \sum_{i=0}^n \frac{1}{i+1} \binom{n}{i}^2$
since $\binom{n}{i} = \binom{n}{n-i}$.

The coefficient of $x^{n+1}$ must be the same on both sides of the equation, so we have
$\frac{1}{n+1} \binom{2n+1}{n+1} = \sum_{i=0}^n \frac{1}{i+1} \binom{n}{i}^2$

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