# Thread: Binomial theorem question check

1. ## Binomial theorem question check

hey guys, just need to check if my working out and solution is correct.

Find the coefficient of $x^6$ in the expansion $(x + 3)^8$

My solution

So: $\sum_{i=0}^{8}$ $\binom{8}{8-i}$ $(x)^{8-i}$ $(3)^i$

now since we need to find coefficient of $x^6$, therefore $i = 6$.

so $\binom{8}{2}$ $(1)^2$ $(3)^6$ = $28 \times 1 \times 739$ = $28 \times 739$

2. Originally Posted by jvignacio
hey guys, just need to check if my working out and solution is correct.

Find the coefficient of $x^6$ in the expansion $(x + 3)^8$

My solution

So: $\sum_{i=0}^{8}$ $\binom{8}{8-i}$ $(x)^{8-i}$ $(3)^i$

now since we need to find coefficient of $x^6$, therefore $i = 6$.

so $\binom{8}{2}$ $(1)^2$ $(3)^6$ = $28 \times 1 \times 739$ = $28 \times 739$
It's not.

The general term is ${8 \choose r} x^{8-r} 3^r$. You require the coefficient corresponding to 8 - r = 6.

3. Originally Posted by mr fantastic
It's not.

The general term is ${8 \choose r} x^{8-r} 3^r$. You require the coefficient corresponding to 8 - r = 6.
So: $\sum_{i=0}^{8}$ $\binom{8}{i}$ $(x)^{8-i}$ $(3)^i$

now since we need to find coefficient of $x^6$, $8 - i = 6$ , so $i = 2.$.

so $\binom{8}{2}$ $(1)^6$ $(3)^2$ = $28 \times 1 \times 9$ = $28 \times 9$ = $252$

this better?