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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.