Show that $\sum_{k=1}^{p}{{p+1}\choose k}2^k=\sum_{k=1}^{p}{{p+1}\choose k}2^{p+1-k}$

Question

Show that $$\sum_{k=1}^{p}{{p+1}\choose k}2^k=\sum_{k=1}^{p}{{p+1}\choose k}2^{p+1-k}.$$

I know that the binomial coefficients are symmetric but how can I proof that the left and the right side are equal?

Answer 1

Define $j=p+1-k$, which covers the same range of values as $k$. Since $\binom{p+1}{k}=\binom{p+1}{j}$, $\sum_k\binom{p+1}{k}2^k=\sum_j\binom{p+1}{j}2^{p+1-j}$.

Answer 2

Hint. Note that $${{p+1}\choose k}={{p+1}\choose p+1-k}$$ then change the summation's index and set $j=p+1-k$.

Answer 3

Use the formula for the binomial sum

• $(a+b)^n =\sum_{k=0}^n \binom{n}{k}a^kb^{n-k}$

and interchange the roles of $a$ and $b$ as follows:

$$(1+2)^{p+1}=1 + \sum_{k=1}^{p}{{p+1}\choose k}2^k +2^{p+1}= 2^{p+1} + \sum_{k=1}^{p}{{p+1}\choose k}2^{p+1-k} + 1 = (2+1)^{p+1}$$