Finding a particular term of a binomial expansion

Question

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?

Answer 1

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 ${7 \choose 2} = {7 \choose 5}$.

Answer 2

$$ (x + y)^n = \sum_{k=0}^{n}{n \choose k}x^ky^{n-k} \quad \Rightarrow \quad (2x^2 + 3)^7 = \sum_{k=0}^{7}{7 \choose k}(2x^2)^{7-k}3^k $$ Like this, $2(7 - k) = 10 \ \Rightarrow \ k =2$. Therefore, the term is ${7 \choose 2}32\cdot 9x^{10}$.

Answer 3

$r$ must be equal to $5$, so the term that will have the exponent $10$ is the $6_{th}$ term..

since to get $r$,

(term desired) - 1

or

$t-1=r$