Binomial theorem question check

Oct 2007
517
6
Santiago
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
Dec 2007
16,948
6,768
Zeitgeist
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.
 
Oct 2007
517
6
Santiago
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?