How to find coefficient of $s^{12}t^{13}$ in the expansion of $(2s-3t)^{25}$

Question

I am trying to find the coefficient of $s^{12}t^{13}$ in the expansion of $(2s-3t)^{25}$.

I don't know if I am doing right, but I know we have to start with $\binom{25}{12}$. When $t^{13}$ is added, becomes more confusing. Any help in handling this problem?

score 0 · Accepted Answer · answered Dec 12 '20 at 11:46

0

The general term is given by $${{25}\choose r}(2s)^r (-3t)^{25-r}$$

You need $r=12$: $${{25}\choose {12}} (2s)^{12} (-3t)^{13} = -{{25}\choose{12}}\cdot 2^{12}\cdot 3^{13} s^{12}t^{13}$$

answered Dec 12 '20 at 11:46

Vishu

14,469

why the value of r is set to 12? – user102859 Dec 12 '20 at 11:50
@user102859 Because you need the power of $s$ to equal $12$, and in my general term, $s$ is raised to the $r$th power. – Vishu Dec 12 '20 at 11:51
I checked an online youtube video: https://www.youtube.com/watch?v=XPmrvRiTruw based on it they use r = 13, not r = 12.... – user102859 Dec 12 '20 at 14:26
@user102859 They must have used a different general expression, either way you’ll get the same answer. – Vishu Dec 12 '20 at 14:28
makes sense. Thanks. – user102859 Dec 12 '20 at 14:29

How to find coefficient of $s^{12}t^{13}$ in the expansion of $(2s-3t)^{25}$

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