How would you prove that 2^x - 2^(x-1) - 2^(x-2)...2^0 = 1 for any x greater than or equal to 1?
T Trenters4325 Junior Member Joined Apr 8, 2006 Messages 122 Jul 2, 2006 #1 How would you prove that 2^x - 2^(x-1) - 2^(x-2)...2^0 = 1 for any x greater than or equal to 1?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Jul 2, 2006 #2 Hello, Trenters4325! Prove that: 2n − 2n−1 − 2n−2 − ⋯ − 20 = 1 \displaystyle \,2^n\,-\,2^{n-1}\,-\,2^{n-2}\,-\,\cdots\,-\,2^0\;=\;1\;\;2n−2n−1−2n−2−⋯−20=1for n ≥ 1\displaystyle n\,\geq\,1n≥1 Click to expand... We have: S = 2n − [1 + 2 + 22 + ⋯ + 2n−1]⏟\displaystyle \,S\;=\;2^n\,-\,\underbrace{\left[1\,+\,2\,+\,2^2\,+\,\cdots\,+\,2^{n-1}\right]}S=2n−[1+2+22+⋯+2n−1] . . . . . . . . . . . . . . . . . . . . . . a geometric series The series has first term a = 1\displaystyle a\,=\,1a=1, common ratio r = 2\displaystyle r\,=\,2r=2, and n\displaystyle nn terms. Its sum is: 1⋅1 − 2n1 − 2 = 2n − 1\displaystyle \,1\cdot\frac{1\,-\,2^n}{1\,-\,2}\;=\;2^n\,-\,11⋅1−21−2n=2n−1 Therefore: \(\displaystyle \:S\;=\;2^n\, -\, \left(2^n\,-\,1)\;=\;1\)
Hello, Trenters4325! Prove that: 2n − 2n−1 − 2n−2 − ⋯ − 20 = 1 \displaystyle \,2^n\,-\,2^{n-1}\,-\,2^{n-2}\,-\,\cdots\,-\,2^0\;=\;1\;\;2n−2n−1−2n−2−⋯−20=1for n ≥ 1\displaystyle n\,\geq\,1n≥1 Click to expand... We have: S = 2n − [1 + 2 + 22 + ⋯ + 2n−1]⏟\displaystyle \,S\;=\;2^n\,-\,\underbrace{\left[1\,+\,2\,+\,2^2\,+\,\cdots\,+\,2^{n-1}\right]}S=2n−[1+2+22+⋯+2n−1] . . . . . . . . . . . . . . . . . . . . . . a geometric series The series has first term a = 1\displaystyle a\,=\,1a=1, common ratio r = 2\displaystyle r\,=\,2r=2, and n\displaystyle nn terms. Its sum is: 1⋅1 − 2n1 − 2 = 2n − 1\displaystyle \,1\cdot\frac{1\,-\,2^n}{1\,-\,2}\;=\;2^n\,-\,11⋅1−21−2n=2n−1 Therefore: \(\displaystyle \:S\;=\;2^n\, -\, \left(2^n\,-\,1)\;=\;1\)
