Let $(a_n)$ be a Fibonacci sequence , Prove that $a_1+a_3+a_5+\cdots+a_{2n-1} = a_{2n}$
In this question $a_n$ is a Fibonacci sequence, I tried solving it with the PMI , problem arose when i was solving it for $P(n)=k+1$ , I could not make both of the sides equal.
Assuming $a_0=0,a_1=1$ we can always add $a_0$ for free, so we get $$ \begin{split} (a_0 + a_1) + a_3 + a_5 + \ldots + a_{2n-1} &= (a_2 + a_3) + a_5 + \ldots + a_{2n-1} \\ &= a_4 + a_5 + \ldots + a_{2n-1} \\ &= a_{2n} \end{split} $$ To make this formal, you can use Mathematical Induction.
$a_1=1;a_2=1;a_{n}=a_{n-1}+a_{n-2};\;n>2$
Induction
for $n=1$ it is true because $a_1=a_2$ by definition
now suppose it is true for $n$
$a_1+a_3+\ldots+a_{2n-1}=a_{2n}$
and let's prove it for $n+1$, that is
$\left(a_1+a_3+\ldots+a_{2n-1}\right)+a_{2n+1}=a_{2n}+a_{2n+1}=a_{2n+2}$
for the definition of the Fibonacci sequence
Hope this helps
