Binomial Theorem

[Dreyfus]

Can someone help me wid da expansion of (1+x)^(-1)

[kangkan]

1+x+x^2+X^3.....

[Noel]

(1+x)^(-1)=1-x+x^2-x^3+x^4....
Use taylor series expansion

[Dreyfus]

Thanks Noel..

[Dreyfus]

Plz help me wid this one...I m stuck...

[Noel]

[Dreyfus]

Sry my fault.....its not (n/k) ....I mistakenly added - b/w k and n ....this is binomial coefficient ie nCk ...

This is the rite one...

[Gayyam]

The given sum equals n(n+1)2^(n-2).
This result may be proved by induction.

[Noel]

is it divergent?

[Noel]

@ameyarane..could you please show the working for this

[Shefali]

Pls tell abt taylor series..

[SINGHAM]

It is quite easy, yaar! Just open the binomial expression, then do some algebra! You should get n*(n+1)*2^{n-2}+n*2^{n-1} which is equal to ameyarane's answer!

[Dreyfus]

Singham...I did the same thing but.....I m not getting (n+1) expression.... can u plz show ur workings...thanks

[Gayyam]

Understanding the LHS, we see that is the number of ways in which we can choose a non-empty subset from a set of n elements, and further choosing two elements from the subset with replacement.
Thus, LHS = summation of nCr*r*r

Now, we can accomplish the same task by first choosing the two elements, and then forming subsets. The number of ways to do that
= n(n-1)...if the elements are distinct, and then 2^(n-2) subsets from the remaining n-2 elements
= n...if the same element is chosen twice, and then 2^(n-1) subsets from the remaining n-1 elements.
Thus, adding -
n*(n-1)*2^(n-2) + n*2^(n-1), we get the desired result.

Hope this helps

[Dreyfus]

Thanks ameyarane

[Dreyfus]

@noel...
I just use expansion after some manipulation I got the RHS in the above image and after that I got the answer n(n-1)2^(n-2)+n2^(n-1)....

[Noel]

Thanks ameyarane and vaibhav :)

[Gayyam]

:) Glad to be of help. Do tell me if you need further explanation. It's kind of a weird approach to prove binomial identities, I know.

[Noel]

Err..as a matter of fact i do..sorry for being a stumbling block

[Gayyam]

Ah. Sure.

Say you are a manager and have a pool of n players in a particular sport of bhoopla. Now, you are to choose a team of r players. Furthermore, you also want to choose a captain and a vice-captain of the team. Bhoopla is a weird sport and you are allowed to have the same player as the captain and the vice-captain. Also, the team can be as small or large as you wish. That is, teams can be as large as n membered, or a one membered.

Before the night of choice, you decide how to attain your objective. The following is your strategy. You decide to first choose the team and then decide who'll be captain and vice-captain. You choose the team of r in nCr ways. Then you choose the captain among the r players in r ways. Similarly, the vice-captain too can be chosen in r ways (since the captain can be the same as the vice-captain). Thus the ways in which you can attain your objective is nCr*r*r. Of course, since you haven't decided on how large the team is, the number of ways you can choose the team is the sum of all such nCr*r*r as r may be any number from 1 to n.

But then, the next day, on the day of choosing, you decide to do things differently. You choose the captain and the vice-captain first, out of the pool of n players, and then the team from the remaining players. Now, if the captain and the vice-captain are different people, then you can choose them in n*(n-1) ways. From the remaining n-2 players, you can choose the team in 2^(n-2) ways. However, if the captain and vice-captain are the same person, then you can choose him in n ways, and then go on to choose the team in 2^(n-1) ways.
Adding them up, we get n*(n-1)*2^(n-2) + n*2^(n-1) ways.

Since both the methods are effectively the same, they must be the same. Thus, sum of (nCr*(r^2)) as r changes from 1 to n = n*(n+1)*2^(n-2).

Do let me know if you need any more explanation. I'll be happy to help. :)