The sum of first `n` odd natural numbers is `2n-1` (b) `2n+1` (c) `n^2` (d) `n^2-1`

Property 6 The sum of first n odd natural numbers is n^(2)

The mean of first n odd natural number is (a) (n+1)/2 (b) n/2 (c) n (d) n^2

The mean of first n odd natural number is (a) (n+1)/2 (b) n/2 (c) n (d) n^2

(i)The sum of the first n natural numbers is [n/3(2n+1)(2n-1),(n^2(n+1)^2)/4,n/6(n+1)(2n+1),(n(n+1))/2]

If the sum of first n even natural numbers is equal to k xx the sum of first n odd natural number then k=(1)/(n) b.(n-1)/(n) c.(n+1)/(2n)d .(n+1)/(n)

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural number then k= a. 1/n b. (n-1)/n c. (n+1)/(2n) d. (n+1)/n

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural number then k= 1/n b. (n-1)/n c. (n+1)/(2n) d. (n+1)/n

The sum of squares of first n natural numbers is given by 1/6 n (n+1)(2n+1) or 1/6 (2n^3 + 3n^2 + n) . Find the sum of squares of the first 10 natural numbers.

Statement 1: The variance of first n even natural numbers is (n^2-1)/4 Statement 2: The sum of first n natural numbers is (n(n+1)/2 and the sum of squares of first n natural numbers is (n(n+1)(2n+1)/6 (1) Statement1 is true, Statement2 is true, Statement2 is a correct explanation for statement1 (2) Statement1 is true, Statement2 is true; Statement2 is not a correct explanation for statement1. (3) Statement1 is true, statement2 is false. (4) Statement1 is false, Statement2 is true

The sum of the first 'n' natural numbers is given by the formula S=(n(n+1))/2 . Find 'n' if the sum is 465.