The value of $$\sum_{r=1}^{15} r^2 \left(\frac{^{15}C_r}{{}^{15}C_{r-1}}\right)$$ is equal to:

We have to evaluate the finite sum

$$\sum_{r=1}^{15} r^{2}\left(\frac{{}^{15}C_{r}}{{}^{15}C_{r-1}}\right).$$

First we simplify the ratio of the two consecutive binomial coefficients. By definition

$${}^{n}C_{k}= \frac{n!}{k!\,(n-k)!}.$$

So, taking $$n=15$$, we can write

$${}^{15}C_{r}= \frac{15!}{r!\,(15-r)!}, \qquad {}^{15}C_{\,r-1}= \frac{15!}{(r-1)!\,\bigl(15-(r-1)\bigr)!} = \frac{15!}{(r-1)!\,(16-r)!}.$$

Now form the required ratio:

$$\frac{{}^{15}C_{r}}{{}^{15}C_{r-1}} =\frac{\dfrac{15!}{r!\,(15-r)!}} {\dfrac{15!}{(r-1)!\,(16-r)!}} =\frac{(r-1)!\,(16-r)!}{r!\,(15-r)!}.$$

Break this fraction into two separate factors:

$$\frac{(r-1)!}{r!}\times\frac{(16-r)!}{(15-r)!}.$$

We know $$r! = r\,(r-1)!$$, hence

$$\frac{(r-1)!}{r!}=\frac{1}{r}.$$

Similarly, $$(16-r)! = (16-r)\,(15-r)!$$, so

$$\frac{(16-r)!}{(15-r)!}=16-r.$$

Multiplying these two results we obtain

$$\frac{{}^{15}C_{r}}{{}^{15}C_{r-1}}=\frac{1}{r}\,(16-r)=\frac{16-r}{r}.$$

Put this back into the original sum. Each term inside the summation becomes

$$r^{2}\left(\frac{16-r}{r}\right)=r^{2}\cdot\frac{16-r}{r}=r\,(16-r).$$

Thus the given sum simplifies to

$$\sum_{r=1}^{15} r\,(16-r).$$

Distribute the product inside the summation:

$$\sum_{r=1}^{15} r\,(16-r)=\sum_{r=1}^{15}\bigl(16r-r^{2}\bigr) =16\sum_{r=1}^{15}r-\sum_{r=1}^{15}r^{2}.$$

Now we evaluate the two standard series that appear.

Formula for the sum of the first $$n$$ natural numbers:

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

Taking $$n=15$$ gives

$$\sum_{r=1}^{15} r=\frac{15\times16}{2}=120.$$

Formula for the sum of the squares of the first $$n$$ natural numbers:

$$\sum_{r=1}^{n} r^{2}= \frac{n(n+1)(2n+1)}{6}.$$

With $$n=15$$ we get

$$\sum_{r=1}^{15} r^{2}= \frac{15\times16\times31}{6}.$$

Compute this step by step:

$$15\times16=240,$$

$$240\times31 = 7440,$$

$$\frac{7440}{6}=1240.$$

Substitute these two results back into our expression:

$$16\sum_{r=1}^{15} r-\sum_{r=1}^{15} r^{2} =16\times120-1240 =1920-1240 =680.$$

So, the entire sum equals

$$680.$$

Hence, the correct answer is Option D.