Anonymous

# What is the Sum? (1+2+...+98+99)?

Is it [n(n + 1)]/2 = [99(100)]/2 = 4950?

If so, how come when I actually wrote out all the pairs, meaning (99 + 1), (98 + 2)...4950 wasn't the answer? I got 5000 when I actually wrote it out because there were 50 pairs, the last being (50 + 50).

Update:

Lol, now I get it. Dumb mistake. Thanks.

Relevance

50 is actually an unpaired number. By adding the pair (50 + 50) twice you added 50 twice, meaning you will get 50 more than the actual answer.

There are 99 numbers between 1 and 99 inclusive, an odd number of numbers, meaning you will have one "loner".

If adding by pairs, what you should have done is (1+99) + (2+98) + (3+97) + .... + (49+51) + 50. Note that this is only 49 pairs, plus 50.

Since it is an AP, so the formula for sumition is :

n/2[2a+(n-1)d]

where n is no of term, here it is 99,

a is the first term which is 1,

d is the comon difference wich is 1,

thus it comes out to be 4950

u r making the pairs, but my friend 50 is the non repeated nomber..which u did actualy thus the answer wil be 49pairs*100+50= 4950

Source(s): Own

There's only one 50, not two.

It doesn't get paired with anything.

1 to 99 = an odd number of numbers,

so 49 pairs (98 numbers) + the number 50 left over.

49 * (100) + 50 = 4950.

(Add in the number 100, and the total is 100 more, or 5050,

50 pairs of numbers which add to 101.)

=(1+99)*99 /2

=50*99

=4950

Your pairs are correct; your application of the formula is wrong. The formula is indeed [n(n + 1)]/2, but you apparently did [n(n - 1)]/2:

100*99/2 = 50*99 = 4950

100*101/2 = 50*101 = 5050

• Anonymous

You got 5000 because the prerogative equals square units as far as net's worth goes, so you need to divide your root digit by the final crucial number and get your answer.

• Anonymous

only 99 numbers = 49.5 pairs

• Anonymous

(1+100)*50=5050