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Question: The sum $\frac{C_1}{C_0}+\frac{2C_2}{C_1}+\frac{3C_3}{C_2}+\dots+\frac{nC_n}{C_{n-1}}$ equals...

The sum C1C0+2C2C1+3C3C2+⋯+nCnCn−1\frac{C_1}{C_0}+\frac{2C_2}{C_1}+\frac{3C_3}{C_2}+\dots+\frac{nC_n}{C_{n-1}} equals

Answer

n(n+1)2\frac{n(n+1)}{2}

Explanation

Solution

The given sum is S=C1C0+2C2C1+3C3C2+⋯+nCnCn−1S = \frac{C_1}{C_0}+\frac{2C_2}{C_1}+\frac{3C_3}{C_2}+\dots+\frac{nC_n}{C_{n-1}}. In summation notation, this is S=∑k=1nkCkCk−1S = \sum_{k=1}^{n} \frac{k C_k}{C_{k-1}}.

Using the definition of binomial coefficients, Ck=(nk)C_k = \binom{n}{k}, the ratio of consecutive coefficients is: CkCk−1=(nk)(nk−1)=n!k!(n−k)!n!(k−1)!(n−k+1)!\frac{C_k}{C_{k-1}} = \frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{\frac{n!}{k!(n-k)!}}{\frac{n!}{(k-1)!(n-k+1)!}} CkCk−1=n!k!(n−k)!×(k−1)!(n−k+1)!n!=(k−1)!k!×(n−k+1)!(n−k)!\frac{C_k}{C_{k-1}} = \frac{n!}{k!(n-k)!} \times \frac{(k-1)!(n-k+1)!}{n!} = \frac{(k-1)!}{k!} \times \frac{(n-k+1)!}{(n-k)!} CkCk−1=1k×(n−k+1)=n−k+1k\frac{C_k}{C_{k-1}} = \frac{1}{k} \times (n-k+1) = \frac{n-k+1}{k}

The kk-th term of the sum is kCkCk−1=k×(n−k+1k)=n−k+1\frac{k C_k}{C_{k-1}} = k \times \left(\frac{n-k+1}{k}\right) = n-k+1.

So, the sum becomes: S=∑k=1n(n−k+1)S = \sum_{k=1}^{n} (n-k+1)

Expanding this sum: For k=1k=1: n−1+1=nn-1+1 = n For k=2k=2: n−2+1=n−1n-2+1 = n-1 For k=3k=3: n−3+1=n−2n-3+1 = n-2 ... For k=nk=n: n−n+1=1n-n+1 = 1

The sum is S=n+(n−1)+(n−2)+⋯+1S = n + (n-1) + (n-2) + \dots + 1. This is the sum of the first nn natural numbers, which is given by the formula n(n+1)2\frac{n(n+1)}{2}.

Therefore, S=n(n+1)2S = \frac{n(n+1)}{2}.