1.

Variance of first n even natural numbers

Answer»

Step-by-step explanation:

FIRST n even natural NUMBERS are 2,4,6,8......2n

Now on calculating the mean we get

μ=2+4+6+8...2nn

=2(1+2+3+4......n)n

=2(n(n+1)2)n=(n+1)

SUM of the SQUARE of the first n even natural number is

∑2nn=2n2=22+42+62.........+(2n)2

=22(12+22+32+........(n)2

We know that 22(12+22+32+........(n)2)=n(n+1)(2n+1)6

=4(n(n+1)(2n+1)6)

=2n(n+1)(2n+1)3

Now calculating the varience we get

=∑2nn=2n2n−μ2

=2n(n+1)(2n+1)3n−(n+1)2

=(n+1)(n−1)3

=(n2−1)3

I hope it had helped u


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