How do I find the term involving $x^{10}$ in the expansion of:
$$
(3+2x^2)^7
$$
I know from the binomial theorem that:
$$
u_{n+1} = {^nC_r a^{n-r}x^r}
$$
and that $n=7, a=3, x=2x^2, r=10$ in this case. But that leads to:
$$
^7C_{10}\cdot3^{-3}\cdot{x^{10}}
$$
Which doesn't match the answer listed ($^7C_{5}\cdot{3^{2}}\cdot{2^5x^{10}}$). Could somebody point me in the right direction?
Best Answer
The binomial theorem for $n=7$ says:
$(a + b)^7 = {7 \choose 0}a^0b^7 + {7 \choose 1}a^1b^6 + {7 \choose 2}a^2b^5 \ldots {7 \choose 7}a^7b^0$
Take $a = 3$, $b = 2x^2$. We need $x^{10}$, so we take the term with ${7 \choose 2}$, which is ${7 \choose 2} \cdot 3^2 \cdot (2x^2)^5$, which is the required answer, as $(2x^2)^5 = 2^5 x^{10}$ and ${7 \choose 2} = {7 \choose 5}$.