0

$$\sum_{i=1}^n n$$

I am confused by the solution to this question: $$\sum_{x=0}^{n-2} 4n=4n(n-1)$$

I know by the distributive property, the 4 can be moved in front of the summation. However, plugging $k=(n-2)$ in $k(k+1)/2$ doesn't return the correct answer and c*k isn't correct either.

What generic form does the summation in the problem take that got that answer?

Thomas Andrews
  • 177,126
Boy
  • 25
  • Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. – Another User Feb 16 '23 at 20:00
  • Sum of a constant (e.g., $n$, which does not depend on the index) is the constant times the number of terms – J. W. Tanner Feb 16 '23 at 20:02
  • 1
    You are confusing $$\sum_{x=0}^{n-2}4n$$ with $$\sum_{x=0}^{n-2}4x.$$ The latter is $$0+4+8+\dots+4(n-2),$$ but the former is $$4n+4n+\dots +4n,$$ where you have $n-1$ terms of the same value. – Thomas Andrews Feb 16 '23 at 20:17
  • 1
    So $$\sum_{i=1}^n i=1+2+\cdots+n=\frac{n(n+1)}2,$$ but $$\sum_{i=1}^n n=n+n+\cdots+n=n^2.$$ – Thomas Andrews Feb 16 '23 at 20:19
  • I see, so it is "n + n", n-2 times?

    I don't know how to embed summations but if this is just the summation of a constant, shouldn't the answer be n(n-2)? Because it's c*n where c = n and n=n-2?

     – Boy Feb 16 '23 at 20:23
  • The summation is starting from 0, not 1. So there is an extra value, and it is $n-1$ and not $n-2$. – Ted Feb 16 '23 at 20:26
  • I see, I was confused because I thought summation from 0 and from 1 were always the same thing. Is that only in the case where the summation is dependent on i? – Boy Feb 16 '23 at 20:29

0 Answers0