Question

Question:

Answer 1

Using:
\[\sum_{r=0}^{n}\left(\begin{matrix}n \\ r\end{matrix}\right)rx^{r}=nx(1+x)^{n-1}\]
Show what:
\[\sum_{r=1}^{n}\left(\begin{matrix}n \\ r\end{matrix}\right)r^{2}=n(n+1)2^{n-2}\]

Answer 2

Show that*

Answer 3

@ParthKohli:
\[(1+x)^{n} = \sum_{r=0}^{n}\left(\begin{matrix}n \\ r\end{matrix}\right)x^{r}\]

Answer 4

Want to learn the proofs Parth?

Answer 5

Yes sir!

Answer 6

Okay, do you know this proof:
\[\left(\begin{matrix}n \\ r\end{matrix}\right)+ \left(\begin{matrix}n \\ r+1\end{matrix}\right)= \left(\begin{matrix}n+1 \\ r+1\end{matrix}\right)\]?

Answer 7

I guess it goes something like:

\( \color{Black}{\Rightarrow \Large {n! \over (n - r)!r!} + {n! \over (n - r - 1)!(r -1)!}}\)

Answer 8

Bingo!
To simplify it; you must know these:
\((r+1)! = (r+1)*r!\)
\((n-r)!= (n-r)(n-r-1)!\)
Are you able to proceed further?

Answer 9

Remember; to find the common factor!

Answer 10

I am afk. Can I talk to you later?

Answer 11

Hang on..it's not right..
This is the correct one.
\[\color{Black}{\Rightarrow \Large {n! \over (n - r)!r!} + {n! \over (n - r - 1)!(r +1)!}}\]

Answer 12

Okay; If I'm still here lol