The sum, $$\sum_{n=1}^{7} \frac{n(n+1)(2n+1)}{4}$$, is equal to

We have to find the value of the finite sum

$$\sum_{n=1}^{7}\frac{n(n+1)(2n+1)}{4}.$$

First we focus on the general term inside the summation. Let

$$T_n=\frac{n(n+1)(2n+1)}{4}.$$

We expand the numerator step by step. We begin with the product inside the brackets:

$$(n+1)(2n+1)=2n^2+3n+1,$$

because

$$2n(n+1)=2n^2+2n \quad\text{and}\quad 1(n+1)=n+1,$$

and adding these two parts gives $$2n^2+2n+n+1=2n^2+3n+1.$$

Now we multiply this result by $$n$$:

$$n\bigl(2n^2+3n+1\bigr)=2n^3+3n^2+n.$$

Hence the general term can be rewritten as

$$T_n=\frac{2n^3+3n^2+n}{4}.$$

Placing this back into the sum, we obtain

$$\sum_{n=1}^{7}T_n=\frac14\sum_{n=1}^{7}\bigl(2n^3+3n^2+n\bigr).$$

We can now separate the summation into three individual sums:

$$\frac14\Bigl(2\sum_{n=1}^{7}n^3+3\sum_{n=1}^{7}n^2+\sum_{n=1}^{7}n\Bigr).$$

To evaluate these, we recall the well-known formulas for sums of powers of the first $$k$$ natural numbers:

$$\sum_{n=1}^{k}n=\frac{k(k+1)}{2},$$

$$\sum_{n=1}^{k}n^2=\frac{k(k+1)(2k+1)}{6},$$

$$\sum_{n=1}^{k}n^3=\Bigl[\frac{k(k+1)}{2}\Bigr]^2.$$

Here, $$k=7$$. Substituting $$k=7$$ into each formula we get

$$\sum_{n=1}^{7}n=\frac{7\cdot8}{2}=28,$$

$$\sum_{n=1}^{7}n^2=\frac{7\cdot8\cdot15}{6}=140,$$

$$\sum_{n=1}^{7}n^3=\Bigl[\frac{7\cdot8}{2}\Bigr]^2=28^2=784.$$

Now we substitute these results back into the expression for the sum:

$$\frac14\Bigl(2\cdot784+3\cdot140+28\Bigr).$$

Calculate each product inside the parentheses:

$$2\cdot784=1568,\quad 3\cdot140=420,$$

so the total inside the brackets is

$$1568+420+28=2016.$$

Finally, dividing by $$4$$ gives

$$\frac{2016}{4}=504.$$

So, the answer is $$504$$.