sandy wrote:
If \(2^k - 2^{k+1} + 2^{k-1} = 2^k\)\(m\), what is the value of \(m\)?
(A) \(-1\)
(B) \(-\frac{1}{2}\)
(C) \(\frac{1}{2}\)
(D) \(1\)
(E) \(2\)
Two ways..
(I) substitute some value for k..
Let k=1, so \(2^k - 2^{k+1} + 2^{k-1} = 2^k\)\(m\), means
\(2^1- 2^{1+1} + 2^{1-1} = 2^1\)\(m......2-4+1=2m.......-1=2m.....m=\frac{-1}{2}\)
(II) Algebraic
\(2^k - 2^{k+1} + 2^{k-1} = 2^k\)\(m\),
\(2^k(1 - 2^1+ 2^{-1} = 2^k\)\(m\),
Equating both sides m=1-2+1/2=-1/2
B