The arithmetic mean of 1- 2- 3- 4- 5- - n is - n-1-2-n-1-2-n-2-n-2-1-

The correct option is B n+1/2 Sum of the first n natural numbers = nbsp;n(n+1)/2 ∴ Mean= ( n(n+1)/2 )/n=n+1/2 ...

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Finding MeanÂ for Grouped Data When Class Intervals Are Not Given

The arithmeti...

Question

The arithmetic mean of 1, 2, 3, 4, 5, ……., n is ____.

$\frac{n + 1}{2}$

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$\frac{n - 1}{2}$

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$\frac{n}{2}$

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$\frac{n}{2} + 1$

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Solution

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Similar questions

n \in N : 1^{2} + 3^{2} + 5^{2} + . . . . . . . + ({2 n - 1)}^{2} = \frac{n (2 n - 1) (2 n + 1)}{3}

1.3.5, 3.5.7, 5.7.9, . . ., (2 n - 1) (2 n + 1) (2 n + 3)

(\forall n \in N)

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 1}

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 2}

1 - 2 n + \frac{2 n (2 n - 1)}{2!} - \frac{2 n (2 n - 1) (2 n - 1)}{3!} + . . . + (- 1)^{n - 1} \frac{2 n (n - 1) . . . (n + 2)}{(n - 1)}

= (- 1)^{n + 1} \frac{(2 n)}{2 (n!)^{2}},

\frac{1^{2}}{(1) (3)} + \frac{2^{2}}{(3) (5)} + . . . . . + \frac{n^{2}}{(2 n - 1) (2 n + 1)} = \frac{n (n + 1)}{2 (2 n + 1)}

\frac{1}{(1) (3)} + \frac{1}{(3) (5)} + . . . . . + \frac{1}{(2 n - 1) (2 n + 1)} = \frac{1}{2 n + 1}

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