Given 1/2 + 1 + 2 + 2^2 + ... + 2^10 = a*2^b + c, find the values of a,b,c.

Consider the Left-Hand-Side (LHS) of the equation first. LHS: 1/2 + 1 + 2 + 2^2 + ... + 2^10. We identify this as a geometric series by noticing that dividing any term u_(n+1) by the preceding term n the result is 2, eg. 1/(1/2) = 2. We also note that there are 12 terms, and that the first term is 1/2. From the formula booklet, section 1.1, we can find an equation for the sum of a finite geometric series: S_n = u_1(1-r^n)/(1-r). Where u_1 is the first term, r is the ratio of successive terms (u_(k+1))/u_k and n is the number of terms. In our case these take the values: u_1 = 1/2 , r = 2 , and n = 12. Substituting these in the equation we have : S_12 = 1/2(1 - 2^12)/(1 - 2) = 1/2(2^12 - 1) = 12^11 - 1/2 We can compare this last result for S_12 with the RHS of the original equation RHS: a2^b + c, to find a = 1, b = 11, and c = -1/2.