Math, asked by adora53, 9 months ago

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​

Answers

Answered by Draxillus
2

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

Answered by pankajroy2
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

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