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Prove that $\sum ^n_{i=1 }(-1)^{i-1}F_{2i}=(-1)^{n+1}F_nF_{n+1}$by induction

my attempt:

for $n=1$ LHS=RHS=1

suppose this is true for $n$

i.,e $F_2+F_4-F_6+F_8-.....+(-1)^{n-1}F_{2n}=(-1)^{n+1}F_nF_{n+1}$

now we have to prove for $n+1$

so consider $F_2+F_4-F_6+F_8-.....+(-1)^{n-1}F_{2n}+(-1)^nF_{2n+2}=(-1)^nF_nF_{n+1}+(-1)^nF_{2n+2}$ ???

how to we processed from here

Understanding math
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1 Answers1

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We shall use induction to prove that

$F_0-F_2+F_4-F_6+\cdots+(-1)^{n}F_{2n} = (-1)^n F_n F_{n+1}$

The base case $N=0$ is easy since both sides are $0$.

For the induction step, we use d'Ocagne's identity: $$ F_{2n}=F_{n+1}^{2}-F_{n-1}^{2}=F_{n}\left(F_{n+1}+F_{n-1}\right) $$ as follows: $$ \begin{align} F_0-F_2+\cdots+(-1)^{n}F_{2n}+(-1)^{n+1}F_{2n+2} &= (-1)^n F_n F_{n+1} + (-1)^{n+1}F_{2n+2} \\ &= (-1)^n (F_n F_{n+1} - F_{2n+2}) \\ &= (-1)^n (F_n F_{n+1} - F_{n+1}(F_{n+2}+F_{n})) \\ &= (-1)^n (-F_{n+1} F_{n+2}) \\ &= (-1)^{n+1} F_{n+1}F_{n+2} \end{align} $$

lhf
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