Proving the binomial identity with n-2 terms

[Lancelot1]

Dear All,

I am trying to prove the following identity:

\[\binom{n}{k}=\binom{n-2}{k}+2\binom{n-2}{k-1}+\binom{n-2}{k-2}\]

My attempt was based on transforming the binomial coefficients into fractions with factorials, and then elimintating similar expressions. Somehow it didn't work out.

I believe that this proof shouldn't be too long. Can you assist ?Thank you in advance.

 

[Evgeny.Makarov]

Using Pascal's identity,
\[
\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}=\left[\binom{n-2}{k-2}+\binom{n-2}{k-1}\right]+\left[\binom{n-2}{k-1}+\binom{n-2}{k}\right]=\binom{n-2}{k-2}+2\binom{n-2}{k-1}+\binom{n-2}{k}.
\]
The identity itself has both algebraic and combinatorial proofs.

 

[Lancelot1]

Thank you !

 

[dromarand]

Please this:
$2^{p-1}\left(2^{p}-1\right)=\sum_{k=0}^{n=\left\lfloor\frac{p-1}{4}\right\rfloor}\binom{2p+2}{4k+p-4n-1}$