renebed
Dec 15, 2010, 04:57 PM
indicated expansion (x+2)^7
galactus
Dec 15, 2010, 05:24 PM
Use the binomial expansion.
There is a pattern to these no matter the exponent.
(a+b)^{7}=a^{7}+7a^{6}b+21a^{5}b^{2}+35a^{4}b^{3}+ 35a^{3}b^{4}+21a^{2}b^{5}+7ab^{6}+b^{7}
See the pattern? The a's powers go down and the b's go up.
The coefficients can be found by multiplying the previous coefficient by the power of a and dividing by one more than the power of b.
i.e. Say we have 7a^{6}b and we want the next coefficient in the line of terms.
\frac{7\cdot 6}{2}=21
Next coefficient: Start with 21a^{5}b^{2}. Then, we have
\frac{21\cdot 5}{3}=35
The first coeffcient is always the leading power. In this case, 7.
See? For this problem, just sub x in for a and 2 in for b into the expansion.