Using mathematical induction prove that
$\displaystyle \sum^{n}_{k=0}\binom{2n+1}{k}=4^n$ for all $n\in\mathbb{N}$
What i try:
Let $$P(n): \sum^{n}_{k=0}\binom{2n+1}{k}=4^n$$
Put $n=1,$ We have $\displaystyle \sum^{1}_{k=0}\binom{3}{k}=4$
Which is true because $\displaystyle \binom{3}{0}+\binom{3}{1}=1+3=4$
Let we assume that $P(n)$ is true for $n=m,$ Then
$$P(m): \sum^{m}_{k=0}\binom{2m+1}{k}=4^m$$
Now i did not understand how can i prove for $n=m+1$. Help me please , Thanks
hint for $k\ge2$: $$\binom{2m+3}{k}=\binom{2m+1}{k}+\binom{2m+1}{k-1} + \binom{2m+1}{k-1}+\binom{2m+1}{k-2}$$ now sum using your induction hypothesis