I am currently studying how to prove the Fibonacci Identity by Simple Induction, shown here, however I do not understand how $-(-1)^n$ becomes $(-1)^{n+1}$. Can anybody explain to me the logic behind this?
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5 Answers
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$-(-1)^n=(-1)^1(-1)^n=(-1)^{n+1}$
answered Sep 17, 2015 at 20:53
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$$-x=(-1)x$$ Now substitute $x=(-1)^n$ $$-(-1)^n=(-1)(-1)^n=(-1)^{n+1}$$
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The negative sign outside the parentheses can be re-written as $-1$: $$-(-1)^n = (-1)(-1)^n$$ That first factor of $(-1)$ can be written as $(-1)^1$, so we have $$(-1)^1(-1)^n$$ and finally the two factors can be combined by adding the exponents: $$(-1)^{n+1}$$
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The powers of $-1$ are $$-1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, \ldots$$ If $n$ is odd, then $(-1)^n = -1$, but if $n$ is even then $(-1)^n = 1$. So if you multiply $(-1)^n$ once by $-1$, it is the same as incrementing $n$. (Technically you can also decrement, but you can't generalize that to other negative numbers).
answered Sep 19, 2015 at 2:41
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Simple there is a $-$ sign in front so you can write $-(-1)^n=(-1)(-1)^n=(-1)^{n+1}$.
answered Sep 19, 2015 at 2:46