formula

At first glance, we see powers of 2 in the denominator of every term except the first one. But if we remember 2^0 = 1, we get our breakthrough. 1/2^0 = 1/1 = 1 Therefore, we stagger the powers of 2 by 1 less than the term we are on: a(n) = 1/2^(n - 1)

Map this out as a function with term number n and value 1, 0 2, 7 3, 14 4, 21 The values jump by 7, but they do so as the n - 1 term. We have the series formula S(n) = 7(n - 1) The problem asks for S(1000) S(1000) = 7(1000 - 1) S(1000) = 7(999) S(1000) = 6,993

If you put $1 a day away and every day you add a dollar to the previous days amount, how much would you have after 100 days Day 1, you have 1 Day 2, you have 1 + 1 = 2 Day 3, you have 1 + 2 = 3 So our formula for day n is: D(n) = n D(100) = 100