Question

find g(1)+g(2)+g(3)...... + g(51) g(x)= 5-x

Answer 1

g(1) = 5-1=4
g(2) = 5-2=3
g(3) = 5-3=2
...
g(51) = 5 - 51 = -46

So you are essentially finding the sum:
4 + 3 + 2 + 1 + 0 - 1 -2 ... - 46

Answer 2

well, i understand that, i was mainly wondering if there was a formula for doing that?

Answer 3

You can also just use summation notation on the formula and do \[\large \sum_{x=1}^{51}(5-x)=\sum_{x=1}^{51}5-\sum_{x=1}^{51}x \]

And use the formula:
\[\large \sum_{x=1}^{n}x=\frac{n(n+1)}{2} \]

Answer 4

ok, so it would be 51(52)/2 ?

Answer 5

yes, for the second summation

Answer 6

the first it would be is 5(6)/2

Answer 7

no not quite.. the formula works if you have the variable "x" in front of the summation, the 5 you have is a constant. When dealing with constants, essentially the formula is telling you you are adding "5" 51-times.

Answer 8

essentially you need the variable you are summing to match with the index of summation:
\[\Large\sum_{\color{red}{x=1}}^{51}\color{red}{x} \], the variables agree, so you can use the above formula. But if you had:
\[\Large\sum_{\color{red}{x=1}}^{51}\color{blue}{a} \], the variables don't agree, so the "a" is constant with respect to the summation index. So in that case you calculate the sum as \(51\cdot a\)

Answer 9

ok, so how would i do that?

Answer 10

in your case the "a" is just 5, so you find the first summation as \(51(5)\)

Answer 11

oh ok i figured out how to do it on my calculator, thank you for your help though'

Answer 12

:)