## anonymous one year ago The sum of K consecutive integers is 41. If the least integer is -40, then k?

1. anonymous

Here's a way of thinking about this. If you're adding consecutive integers and the first is -41, then you're adding:$-41+(-40)+(-39)+(-38)+...=41$However, at some point, you're going to get into the positive numbers when you're adding.$-41+(-40)+(-39)+...+(-3)+(-2)+(-1)+(0)+(1)+(2)+(3)+...=41$At the point where you get to the positive numbers, the negatives and positives will begin to cancel out. This is because -3 + 3 = 0, -2 + 2 = 0, -1 + 1 = 0. Does this help you with the rest of the problem?

2. anonymous

The cutoff part says $$+...=41$$

3. phi

the least number is -40

4. misty1212

oops

5. anonymous

Ok, if the first numebr is -40, just omit the -41 in my equations.

6. dan815

does that equation work

7. anonymous

hmmmmm so it's 41 then?

8. anonymous

Not quite. 41 would be the last term that you would be adding to get the sum to be 41. However, k is the number of numbers you're adding.

9. anonymous

ohhhhh crap .... it's 81

10. anonymous

-40 to 0 is 41 numbers. 1 to 41 is 41 numbers. So, 41 + 41 should be what k is. Is this making sense to you?

11. freckles

$\sum_{i=1}^k(-40+i)=41 \\ -40k+\frac{k(k+1)}{2}=41$

12. freckles

13. anonymous

hmmm you know what i do is i add 41 to -40 and then multiply by 81

14. freckles

I think I should have started it at i=0

15. anonymous

The key here is that they are "consecutive" integers. That's why just multiplying like that doesn't work.

16. freckles

$-40(k+1)+\frac{k(k+1)}{2}=41$

17. anonymous

crap so it's 82

18. freckles

almost

19. anonymous

huh?

20. anonymous

@freckles I got 82 as well. Might've had an error though.

21. freckles

you said 81 earlier that is right

22. freckles

that is what you get when you solve that quadratic equation above $\sum_{i=0}^k (-40+i)$ this does mean -40+(-39)+(-38)+...+(-40+k) or if you used those formula things you can write that sum thingy as $-40(k+1)+\frac{k(k+1)}{2} \text{ which we want to this output } 41 \\ \text{ so we need \to solve } \\ -40(k+1)+\frac{k(k+1)}{2}=41$

23. freckles

$\frac{-80(k+1)}{2}+\frac{k(k+1)}{2}=41 \\ \frac{(k+1)}{2}(-80+k)=41 \\ (k+1)(-80+k)=82 \\ k^2-79k-80=82 \\ k^2-79k-80-82=0 \\ k^2-79k-162=0 \\ (k+2)(k-81)=0$

24. freckles

only one of these k's make sense

25. dan815

82 is right i believe

26. dan815

|dw:1436823495199:dw|

27. anonymous

@freckles You can use the sum of an arithmetic series.$a_k=-40, -39, ...$$S_k=\frac{k}{2}(2a_1+(k-1)d)$$41=\frac{k}{2}(k-81)$$82=k^2-81k$$0=k^2-81k-82$$0=(k-82)(k+1)$$k=-1 ~or ~82$And -1 is extraneous.

28. anonymous

If you prefer the algebraic method.

29. freckles

I gave an algebraic method above 81 works

30. freckles
31. anonymous

By starting at i = 0, the number of terms in that series is actually k + 1. I think that's the error.

32. anonymous

thanks everyone.

33. dan815

yeah 82 is right

34. freckles

then why is wolfram saying 81 that is weird

35. dan815

freckles the indice starts at 0

36. dan815

not 1

37. anonymous

The number of terms can easily be calculated by b-a+1 given$\sum_{a}^{b}$What you used equates to k+1 terms.

38. dan815

|dw:1436823854020:dw|

39. freckles

so I should have used this: $\sum_{i=1}^{k}(-40+(i-1)) =41 \\ -40k+\frac{k(k+1)}{2}-k =41$ i think this should arrive to the correct answer and yes this is k terms now my bad

40. freckles

$-80k+k(k+1)-2k=82 \\ k^2+k-82k=82 \\ k^2-81k-82=0 \\ (k-82)(k+1)=0$