Answered

Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

1+2-3+4+5-6+7+8-9...+97+98-99

Sagot :

sqdancefan

Answer:

  1584

Step-by-step explanation:

The sum of this sequence can be found a number of ways. One way is to recast it as the series whose terms are groups of three terms of the given series.

__

series of partial sums

The partial sums, taken 3 terms at a time, are

  1+2-3 = 0

  4+5-6 = 3

  7+8-9 = 6

...

  97+98-99 = 96

So the original series is equivalent to ...

  0 +3 +6 +... +96 = 3×1 +3×2 +... +3×32 = 3×(1 +2 +... +32)

That is, the sum is 3 times the sum of the consecutive integers 1..32.

__

consecutive integers

The sum of integers 1..n is given by the equation ...

  s(n) = n(n+1)/2

__

series sum

Using this to find the sum of our series, we find it to be ...

  series sum = 3 × (32)(33)/2 = 1584

_____

Alternate solution

The given series is the sum of integers 1-99, with 6 times the sum of integers 1-33 subtracted. That is, ...

  1 + 2 - 3 + 4 + 5 - 6 = 1+2+3+4+5+6 -2(3 +6) = 1+2+3+4+5+6 -6(1+2)

Continuing on to ...97 +98 -99 gives the result s(99) -6s(33).

Computed that way, we find the sum to be ...

  (99)(100)/2 -6(33)(34)/2 = 4950 -3366 = 1584

We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.