SOLUTION: Find the expansion of (2+x)^5, giving your answer in ascending powers of x, and by letting x=0.01 or otherwise, find the exact value of 2.01^5.

Question 718110: Find the expansion of (2+x)^5, giving your answer in ascending powers of x, and by letting x=0.01 or otherwise, find the exact value of 2.01^5.

Found 2 solutions by jsmallt9, lynnlo:
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
To find the expansion you have two choices:
Multiply (2+x)(2+x)(2+x)(2+x)(2+x) by hand; or...
Use knowledge about binomial expansions to go (almost) directly to the expansion. The expansion of will take the form of:

where the a's are the coefficients. (Note how the exponents of 2 and x in each term adds up to 5 (the exponent to which (2+x) is being raised). If we were raising (2+x) to the 11th power the two exponents of each term would add up to 11.) The hard part is figuring out the a's. For these you can either use
Pascal's triangle; or
the part of the Binomial Theorem formula for the coefficients:

where n is the power to which the binomial is being raised (in this case 5)
where p and q are the exponents on the individual factors of that term (in this case the exponents on 2 and x)
and z! is read "z factorial" and means 1 * 2 * 3 * ... * z
For example:

So

Once you have the expansion worked out, then we will substitute in 0.01 in for x. Hints:
Since and since raising to various powers is probably easier than raising 0.01 to the various powers.; or...
Use the Remainder Theorem. If you use synthetic division to divide the expansion by (x-0.01) then the remainder will be the number you are looking for.

Answer by lynnlo(4176) (Show Source): You can put this solution on YOUR website!
x^5+10x^4+40x^3+80x^2+80x+32
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