Question 718110
To find the expansion you have two choices:<ul><li>Multiply (2+x)(2+x)(2+x)(2+x)(2+x) by hand; or...</li><li>Use knowledge about binomial expansions to go (almost) directly to the expansion. The expansion of {{{(2+x)^5}}} will take the form of:
{{{a[0]2^5x^0+a[1]2^4x^1+a[2]2^3x^2+a[3]2^2x^3+a[4]2^1x^4+a[5]2^0x^5}}}
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<ul><li>Pascal's triangle; or</li><li>the part of the Binomial Theorem formula for the coefficients:
{{{n!/(p!*q!)}}}
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:
{{{a[3] = 5!/(2!*3!) = (1*2*3*4*5)/((1*2)*(1*2*3))  = (cross(1*2*3)*4*5)/((1*2)*(cross(1*2*3))) = (4*5)/(1*2) = 10}}}
So {{{a[3]2^2x^3 = 10*4*x^3 = 40x^3}}}</li></ul></li></ul>
Once you have the expansion worked out, then we will substitute in 0.01 in for x. Hints:<ul><li>Since {{{0.01 = 10^((-2))}}} and since raising {{{10^((-2))}}} to various powers is probably easier than raising 0.01 to the various powers.; or...</li><li>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.</li></ul>