Question

use binomial theorem to expand (1+x)^7

Answer 1

@vishweshshrimali5

Answer 2

Okay lets see:

\[\large{(1+x)^n = ^n C_{0} x^n + ^nC_1 x^{n-1} + ... +^nC_n}\]

Answer 3

Or I can write it as:

\[\large{(1+x)^n = \sum_{i=1}^{n+1} \ ^nC_{i-1}x^{n-i-1}}\]
\[\large{\implies (1+x)^n = \sum_{i=0}^{n} \ ^nC_{i} x^{n-i}}\]

Answer 4

Here your n is 7

Answer 5

ok..

Answer 6

\[(a+b)^n=\sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right)\]=\[a ^{n-k}b ^{k}\]\[\left(\begin{matrix}n \\ k\end{matrix}\right)=n _{C}k=n!/(n-k)!k!\] the product of all the whole numbers between 1 and n", so, for instance \[7!/(7-1)!1!=1*2*3*4*5*6*7/1*2*3*4*5*6*1=7 \] 7 would be the coeffcient.
\[(1+x)^7\]=\[7C0 (1^7)+(7C1)(1^6*x)+7C2(1^5*x^2)+7C3(1^4*x^3)+7C4(1^3*x^4)+7C5(1^2*x^5)+7C6(1*x^6)+7C7x^7\]

Answer 7

I don't really understand..

Answer 8

The given answer is
1+7x+21x^2+35x^3+35x^4+21x^5+7x^6+x^7

Answer 9

This is how I would approach the problem :)
One sec...

Answer 10

\[\large\rm (1+x)^7=\]\[\rm \text{_}(1)^7(x)^0+\text{_}(1)^6(x)^1+\text{_}(1)^5(x)^2+\text{_}(1)^4(x)^3+\text{_}(1)^3(x)^4+\text{_}(1)^2(x)^5+\text{_}(1)^1(x)^6+\text{_}(1)^0(x)^7\]
Each term has the 1 and the x.
The power starts at 7 on the 1, and count down to 0.
They start at 0 on the x, and count up to 7.

So we have 8 terms.
I left a little space in front for the coefficients.
You can calculate the coefficients the way that vish and abmon have shown you.
But I prefer to just use Pascal's Triangle. :d

The 8th row will correspond to the coefficients of a 7th degree binomial.