Question

im getting lost trying to solve this, could you help clarify

Answer 1

The rule for the sum of integers elevated to the second power is the following:

$\sum ^n_{k\mathop=1}k^2=\frac{n(n+1)(2n+1)}{6}$

The series given can be divided in the following way:

$\sum ^{25}_{k\mathop=5}k^2=\sum ^{25}_{k\mathop=1}k^2-\sum ^4_{k\mathop=1}k^2$

That is, to the sum of the number of k squared from 1 to 25 we subtract the sum of the numbers of k squared from 1 to 4. Now we can use the rule for the sum of integers to find each sum.

$\sum ^{25}_{k\mathop=1}k^2=\frac{25(25+1)(2(25)+1)}{6}$

Solving the operations:

$\sum ^{25}_{k\mathop=1}k^2=5525$

Using the formula for the next sum:

$\sum ^4_{k\mathop=1}k^2=\frac{4(4+1)(2(4)+1)}{6}$

Solving the operations:

$\sum ^4_{k\mathop=1}k^2=30$

Replacing in the formula for the given sum, we get:

$\sum ^{25}_{k\mathop=5}k^2=5525-30=5495$

If we were asked for the sum from 3 to 5, we could do the following:

$1^2+2^2+3^2+4^2+5^2-(1^2+2^2)=3^2+4^2+5^2$