The sum #1 + 3 + 7 + 15 + 31 ...# to #100# terms is?

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Answer 1 · 2018-02-15T14:02:25

See below.

Calling

#S_n = sum_(k=0)^n a_k# with

#a_k = 2a_(k-1) +1# we have

#a_k = 2^(k+1)-1# and

#S_n = sum_(k=0)^n (2^(k+1)-1) = -(n+1)+2sum_(k=0)^n 2^k# or

#S_n = 2 ((2^(n+1)-1)/(2-1)) -(n+1) = 2 (2^(n+1)-1)-(n+1)# and then

#S_100 = 2 (2^101 - 1) - 101 = 5070602400912917605986812821401#