f(n+1)=2f(n)-1

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Solution for f(n+1)=2f(n)-1 equation:


Simplifying
f(n + 1) = 2f(n) + -1

Reorder the terms:
f(1 + n) = 2f(n) + -1
(1 * f + n * f) = 2f(n) + -1
(1f + fn) = 2f(n) + -1

Multiply f * n
1f + fn = 2fn + -1

Reorder the terms:
1f + fn = -1 + 2fn

Solving
1f + fn = -1 + 2fn

Solving for variable 'f'.

Move all terms containing f to the left, all other terms to the right.

Add '-2fn' to each side of the equation.
1f + fn + -2fn = -1 + 2fn + -2fn

Combine like terms: fn + -2fn = -1fn
1f + -1fn = -1 + 2fn + -2fn

Combine like terms: 2fn + -2fn = 0
1f + -1fn = -1 + 0
1f + -1fn = -1

Reorder the terms:
1 + 1f + -1fn = -1 + 1

Combine like terms: -1 + 1 = 0
1 + 1f + -1fn = 0

The solution to this equation could not be determined.

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