Alternate method for the summation of the given binomial sum

Question

If $c_{0}, c_{1}, c_{2}, \ldots . . c_{15}$ are the binomial coefficients in the expansion of $(1+x)^{15}$, then the value of $\frac{c_{1}}{c_{0}}+2 \frac{c_{2}}{c_{1}}+3 \frac{c_{3}}{c_{2}}+\ldots+15 \frac{c_{15}}{c_{14}}$ is.

At first I was trying to get it from differentiation but that gives only the numerator , no terms of th binomial coefficients in the denominator , then I check in general what each term represents : $r.\frac{\binom{n}{r}}{\binom{n}{r-1}}$ , from that I got the pattern and solved it , but I would like to know if there exists a method which uses diff/integration to get the required sum .

Answer 1

Yes, the computation can be derived via differentiation, in a convoluted manner.

Let $f(x) = (1 + x)^{15}.$

For $k \in \{0,1,2,\cdots,15\}$,
let $f^{(k)}(x)$ denote the $k$-th derivative of $f(x)$.
Let $f^{(k)}(0)$ denote the $k$-th derivative of $f(x)$, evaluated at $x = 0.$

Then

$$f^{(k)}(0) = \frac{(15)!}{(15-k)!} = k! \times \binom{15}{k} = k! \times C_k.$$

Therefore, the desired summation is equivalent to

$$\sum_{k=1}^{15} \frac{f^{(k)}(0)}{f^{(k-1)}(0)}. \tag1 $$

(1) above, which is the desired computation, can be alternatively expressed.

Since $f(x) = (1 + x)^{(15)}$, you have that

$\displaystyle f^{(k)}(0) = \frac{(15)!}{(15 - k)!}.$

Therefore,

$$\frac{f^{(k)}(0)}{f^{(k-1)}(0)} = 15 - (k-1) = 16 - k.\tag2 $$

Therefore, (1) above can be alternatively expressed as

$$\sum_{k=1}^{15} (16 - k)$$

$$= [16 \times 15] - \frac{16 \times 15}{2} = \frac{16 \times 15}{2}.$$

Answer 2

$$ \frac{r\cdot\binom{n}{r}}{\binom{n}{r-1}} = \frac {r\cdot n! \cdot (r-1)!\cdot (n-r+1)!} {n! \cdot r!\cdot (n-r)!} = \frac {n! \cdot r!\cdot (n-r+1)!} {n! \cdot r!\cdot (n-r)!} = $$ $$ =\frac {n! \cdot r!\cdot (n-r)! \cdot (n-r+1)} {n! \cdot r!\cdot (n-r)!} = n-r+1$$

$$ \sum_{r=1}^{r=15} \frac{r\cdot\binom{15}{r}}{\binom{15}{r-1}} = 15+14+13+...+1 = 8 \cdot 15 = 120 $$

General:

$$ \sum_{r=1}^{r=n} \frac{r\cdot\binom{n}{r}}{\binom{n}{r-1}} = n+(n-1)+(n-2)+...+1 = \frac{(n+1)\cdot n}{2} = \binom{n+1}{2} $$