# Binomial theorem question check

• May 19th 2010, 12:15 AM
jvignacio
Binomial theorem question check
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 $\displaystyle \binom{8}{2}$ $\displaystyle (1)^2$ $\displaystyle (3)^6$ = $\displaystyle 28 \times 1 \times 739$ = $\displaystyle 28 \times 739$
• May 19th 2010, 12:23 AM
mr fantastic
Quote:

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

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.
• May 19th 2010, 12:32 AM
jvignacio
Quote:

Originally Posted by mr fantastic
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?