Question

The arithmetic mean of the set of observations 1²,2²'3² ........ n²?is
(n+1)(2n+1)
n(n+1)
(n-1)(2n +1)
D (n+1)(2n-1)
D) -
96
6​

Answer 1

Heya,

To answer this question

we need to summarise some formulas .They are mentioned below

$1 + 2 + 3 + 4 + ....n = \frac{n(n + 1)}{2} \\ \\ {1}^{2} + {2}^{2} + {3}^{2} + ..... {n}^{2} = \frac{1n(n + 1)(2n + 1)}{6} \\ \\ {1}^{3} + {2}^{3} + {3}^{3} + ....... {n}^{3} = { \frac{n(n + 1)}{ {2}^{2} } }^{2}$

Hence, mean whose formula is

$\frac{sum \: of \: all \: obs}{no \: of \: obs}$

Applying the above formula, we get mean =

$\frac{ \frac{1}{6} n(n + 1)(2n + 1)}{n}$

Hence, the required mean is :-

$\frac{1(n + 1)(2n + 1)}{6}$

Regards

Kshitij

Answer 2

Answer:

here is ur answer

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

2

n(n+1)

1

2

+2

2

+3

2

+.....n

2

=

6

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

1

3

+2

3

+3

3

+.......n

3

=

2

2

n(n+1)

2

Hence, mean whose formula is

sumofallobsnoofobs\frac{sum \: of \: all \: obs}{no \: of \: obs}

noofobs

sumofallobs

Applying the above formula, we get mean =

16n(n+1)(2n+1)n\frac{ \frac{1}{6} n(n + 1)(2n + 1)}{n}

n

6

1

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

Hence, the required mean is :-

1(n+1)(2n+1)6\frac{1(n + 1)(2n + 1)}{6}

6

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

here