Don't just add!
Find the sum of all positive integer values <100.
The answer is 4950.
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3 solutions
We need to calculate, 1 + 2 + 3 + … + 9 9 This is an arithmetic progression with a = 1 , d = 1 , n = 9 9 and l = 9 9 S u m = 2 n ( a + l ) = 2 9 9 ⋅ ( 9 9 + 1 ) = 9 9 × 5 0 = 4 9 5 0
| 1 | + 2 | + 3 | ........ | + 99 |
| 99 | +98 | +97 | ........ | +1 |
If you pair these numbers up the sum of each pair is 100. There will be 99 pairs of numbers so 99 x 100 = 9900. However you have doubled the original list, so now you must half the sum 9900/2 = 4950
I really liked the fact that you showed the most elementary method. But let's generalize:
Assume the positive integers are from 1 to n , then the sum looks like
S = 1 S = n + + 2 ( n − 1 ) + + 3 ( n − 2 ) + + ⋯ ⋯ + + ( n − 1 ) 2 + + n 1
Summing up both we get
2 S = ( n + 1 ) + ( n + 1 ) + ( n + 1 ) + ⋯ n terms = n ( n + 1 )
Thus the sum is
S = 2 n ( n + 1 )
An very famous elementary result!
Follow up problem:
a) Find the sum of squares of all positive integers values < 1 0 0 .
b) Generalize for the sum of squares of naturals from 1 to n (both inclusive).
summation of all the integers less than 100 is-=(1+2+3+4+....................99)
so, the rule is = n × ( n + 1 ) ÷ 2