- #1
sleepingMantis
- 25
- 5
Hello,
I am working through Spivak for self study and sharpening my math skills. I have become stuck on an exercise.
What I need to show is the following:
$$
(a + b) \sum_{j = 0}^{n} \binom nj a^{n-j}b^{j} = \sum_{j = 0}^{n + 1} \binom{n+1}{j} a^{n-j + 1}b^{j}
$$
My attempt, starting from the LHS. Expanding the brackets:
$$
(a + b) \sum_{j = 0}^{n} \binom nj a^{n-j}b^{j} = a \sum_{j = 0}^{n} \binom nj a^{n-j}b^{j} + b \sum_{j = 0}^{n} \binom nj a^{n-j}b^{j}
\\ = \sum_{j = 0}^{n} \binom nj a^{n+1-j}b^{j} + \sum_{j = 0}^{n} \binom nj a^{n-j}b^{j+1}
$$
Now I know a useful equality (that I proved earlier) is:
$$
\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}
$$
However I am not sure how to proceed. Any tips would be much appreciated!
I am working through Spivak for self study and sharpening my math skills. I have become stuck on an exercise.
What I need to show is the following:
$$
(a + b) \sum_{j = 0}^{n} \binom nj a^{n-j}b^{j} = \sum_{j = 0}^{n + 1} \binom{n+1}{j} a^{n-j + 1}b^{j}
$$
My attempt, starting from the LHS. Expanding the brackets:
$$
(a + b) \sum_{j = 0}^{n} \binom nj a^{n-j}b^{j} = a \sum_{j = 0}^{n} \binom nj a^{n-j}b^{j} + b \sum_{j = 0}^{n} \binom nj a^{n-j}b^{j}
\\ = \sum_{j = 0}^{n} \binom nj a^{n+1-j}b^{j} + \sum_{j = 0}^{n} \binom nj a^{n-j}b^{j+1}
$$
Now I know a useful equality (that I proved earlier) is:
$$
\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}
$$
However I am not sure how to proceed. Any tips would be much appreciated!
Last edited: