# Binomial theorem question check

#### jvignacio

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

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

My solution

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

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

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

#### mr fantastic

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

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

My solution

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

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

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

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

#### jvignacio

It's not.

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

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

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

this better?