How do you use the binomial series to expand # (2 + x)^11 #?

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Answer 1 · 2015-12-06T12:01:05

Multiply a row of Pascal's triangle by a sequence of descending powers of #2# to find:

#(2+x)^11=2048 + 11264x + 28160x^2 + 42240x^3 + 42240x^4 + 29548x^5 + 14784x^6 + 5280x^7 + 1320x^8 + 220x^9 + 22x^10 + x^11#

#(2+x)^11 = sum_(k=0)^11 ((11),(k)) 2^(11-k)x^k#

We can find the values of #((11),(k))# from Pascal's triangle:
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Write out the row beginning #1#, #11#:

#1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1#

Write out the powers of #2# in descending order from #2^11# to #2^0#:

#2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1#

Multiply these two sequences together to get:

#2048, 11264, 28160, 42240, 42240, 29548, 14784, 5280, 1320, 220, 22, 1#

These are the coefficients to find:

#(2+x)^11=2048 + 11264x + 28160x^2 + 42240x^3 + 42240x^4 + 29548x^5 + 14784x^6 + 5280x^7 + 1320x^8 + 220x^9 + 22x^10 + x^11#