Binomial Theoreom: Find 1st 3 terms in expansion of (2+u)^5; Find...

[cired2002]

(i) Find the first three terms, in ascending powers of u in the expansion of (2+u)^5.
(ii) By replacing u with 2x-5x^2, find the coefficient of x^2 in the expansion of (2+2x-5x^2)^5

I have done (i) 32+80u+80u^2. But I couldn't do part (ii). Is there any simpler way to do (ii) now that i found (ii) other than expanding it just like that?

 

[tkhunny]

Are you doing the substitution before or after ^5?
There may be a reason why Part (i) was done first.

 

[cired2002]

Are you doing the substitution before or after ^5?
There may be a reason why Part (i) was done first.

So we have to expand all of it out?
I only expanded until u²

 

[JeffM]

\(\displaystyle v = 2 + u \implies v^m = \displaystyle \sum_{j=0}^m \dbinom{m}{j} 2^{(m-j)}u^j.\)

\(\displaystyle \therefore v^5 = 32 + 80u + 80u^2 + \text {terms in higher powers of } u.\)

You got this far. Good.

\(\displaystyle u = (2x - 5x^2) \implies u^j = \displaystyle \sum_{k=0}^j \dbinom{j}{k} (2x)^{(j-k)}(-\ 5x^2)^k = \sum_{k=0}^j \dbinom{j}{k} (2)^{(j-k)}(-\ 5)^k(x)^{(j-k+2k)} = \)

\(\displaystyle \displaystyle x^j * \sum_{k=0}^j \dbinom{j}{k} (2)^{(j-k)}(-\ 5x)^k.\)

Thus, powers of u higher than 2 will contain no terms in x squared. You just have to figure out things in terms of u and u squared. Clear now?