document.write( "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.
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Algebra.Com's Answer #440765 by jsmallt9(3758) $\"\"$ $\"About$

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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:
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\n" ); document.write( "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:
  \n" ); document.write( " $\"n%21%2F%28p%21%2Aq%21%29\"$
  \n" ); document.write( "where n is the power to which the binomial is being raised (in this case 5)
  \n" ); document.write( "where p and q are the exponents on the individual factors of that term (in this case the exponents on 2 and x)
  \n" ); document.write( "and z! is read \"z factorial\" and means 1 * 2 * 3 * ... * z
  \n" ); document.write( "For example:
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  \n" ); document.write( "So $\"a%5B3%5D2%5E2x%5E3+=+10%2A4%2Ax%5E3+=+40x%5E3\"$

\n" ); document.write( "Once you have the expansion worked out, then we will substitute in 0.01 in for x. Hints:

Since $\"0.01+=+10%5E%28%28-2%29%29\"$ and since raising $\"10%5E%28%28-2%29%29\"$ 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.

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