How to evaluate $\sum_{ r=1}^{16}(5r-7)$?

Question

I'm self-studying from Stroud & Booth's "Engineering Mathematics" and in the "Binomials" chapter, one of the last exercises is to evaluate:

$$\sum_{ r=1}^{16}(5r-7)$$

This has got me confused, as all the examples used the summation going up to $n$, but never to a number (16) in this case.

Can anyone help me out?

Answer 1

$$\sum_{r=1}^{16}(5r-7) = 5\sum_{r=1}^{16} r - 7\sum_{r=1}^{16} 1 = 5\sum_{r=1}^{16} r - 7\cdot 16$$

Now use the fact that $$\sum_{r=1}^{n}r=\frac{n(n+1)}{2}$$ to calculate what $\sum_{r=1}^{16} r$ is equal to (that is, just replace $n$ with $16$).

Answer 2

This is an arithmetic series, which is a series where the difference between successive terms is constant. You should be able to see that the starting term is $a=-2$ and there are $n=16$ terms, and the common difference is $5$. So the sum is $$\frac{n}{2}(2a+(n-1)d) = \frac{16}{2}\times (2\times(-2)+(16-1)\times 5)=568.$$

You can see Wikipedia for a proof that an arithmetic series with $n$ terms, starting term $a$ and common difference $d$ has sum $\frac{n}{2}(2a+(n-1)d)$.

Answer 3

You have $$ 5(1+2+3+...+16) -16(7)$$

$$=5(8)(17)-16(7)=568$$