Question

find the coefficient of x^3 in the expansion of (1+2x)^6
can you help me please
AM USING THIS FORMULA nCr a^r b^n-r

Answer 1

You can simply expand this using binomial theorem, or just expanding the brackets manually.

I'm not overly familiar with that formula; although I know how to use it. Let me see if I can get it.

Answer 2

ok

Answer 3

If you expand (1+2x)^6 you get:
\[64 x^6+192 x^5+240 x^4+160 x^3+60 x^2+12 x+1\]

Using this formula seems a weird method when you can simply expand the brackets using binomial theorem.

So the coefficient on the x^3 is 160.

Answer 4

The point of the formula is that you won't always want to expand the expression in full.

For example here it was still more work to expand the entire expression as chaise has done rather rather than just calculate the one term in which you are interested.

In general,
\[jth \ term \ of \ (a+b)^n = {}^nC_{j-1}a^{n-j+1}b^{j-1}\]
Here n = 6, a = 1, b = 2x and j = 4
\[{}^nC_{j-1} = {}^6C_3 = \frac{6!}{3! 3!} = 20\]
and
\[ a^{n-j+1}b^{j-1} = 1^4(2x)^3 = 8x^3\]
thus the 4th term in total is
\[20 . 1 . 8x^3 = 160x^3\]

This might seem like a lot of work, but is it really more than calculating the entire binomial theorem? I don't think so.

But if you're still not convinced, what if n = 10, or n = 20?

So it's worth learning this method because you won't always want to calculate the complete expansion.

Answer 5

I agree with James-san... I've encountered questions that required me to find certain terms and the expression was to the 15th or 17th power (can't remember)... It'll be a pain in the neck if you expand it with binomial. So I just used James-san's method...