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Sum of First N Odd Consecutive IntegersGmatKiss wrote:The below compilation may help!!
Sum of all consecutive integers:
N (N+1) / 2 ------- (I)
where N is the number of terms.
Example: 1+2+3+........ + 100
N=100
(I) => 100 (101) / 2
=> 50 (101) = 5050
First N consecutive Integers
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SUM OF EVEN NUMBERS:
Formula: N(N+1)
How to Find N = (First Even + Last Even)/2 - 1
Example: 2+4+6+ ....... 100
N = (2+100)/2 - 1 = 50
Sum of first 50 positive even integers = (50)(51) = 2550
Sum of First N Even Consecutive Integers
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SUM OF ODD NUMBERS:
If N = number of odd terms then sum = (N)^2
Sum of squares of 1st n consecutive natural nos = n(n+1)(2n+1) / 6
Sum of cubes of the 1st n consecutive natural nos. = [n(n+1)/2]^2
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