Can someone help me with this?

Question

Find the sum of the first 200 positive integers

Answer 1

(1+200)*200/2
=201*100
=20100

Answer 2

It is easy there is a formula and it is obtained like this. Let S be the sum of the n subsequent integers, so:

S= 1 + 2 + ......+ (n-2) + (n-1) + n

but, since the order in which the sum is performed does not affect the result, it can also be put in the following way

S= n + (n-1) + (n-2) + ...... + 2 + 1

now we combine this two expressions like this:

S= 1 + 2 + 3 +......+ (n-2) + (n-1) + n
+ S= n + (n-1) + (n-2) + ...... + 3 + 2 + 1
___________________________
2S= (n+1) + (n-1+2) + (n-2+3).....+ (n-2+3) + (n-1+2) + (n+1)

and since there are n terms in this sum (the sum runs over the n first integers), thus

2S=n(n+1)

which in turn implies that S=n(n+1)/2.

So for n=200 this results in S=200(201)/2=20100

I hope this may help you. Bye.

Answer 3

An integer is a whole number. So basically, I believed you are being asked to add up:

1 + 2 + 3 + 4 ........+ 200

There is a formula for this:

Sum = n(a1 + an)/2

Where n = difference between each term in the progression - in this case, 1
a1 = first term in progression (1)
an = nth term in progression (200)

sum = 200(1+200)/2 = 40200/2 = 20100

Edit: why would someone give this a thumbs down? The answer is accurate AND it has a reference! Some people are just mean I guess :-)

Answer 4

(1+200) x 200/2
= 201 x 100
= 20100

Answer 5

100

Answer 6

(200)(201)/2 = 100(201) = 20100.

Answer 7

Sn = a + (a + d) + (a + 2d) + -----(a + (n - 1).d)
Sn = (a + (n - 1)d) + ----------(a + d) + a
2Sn = n.(2a + (n - 1).d)
Sn = (n / 2).[ 2a + (n - 1).d ]
S200 = 100 . (2 + 199 x 1)
S200 = 100 x 201
S200 = 20100