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2 years ago

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

Mathematics

1 answer:

zaharov [31]2 years ago

3 0

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.

<h3>series of partial sums</h3>

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.

<h3>consecutive integers</h3>

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

s(n) = n(n+1)/2

<h3>series sum</h3>

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

series sum = 3 × (32)(33)/2 = 1584

_____

<em>Alternate solution</em>

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

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