The sum of the first n even natural numbers is .....

n^{2}

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n^{2} + 2

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

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n (n + 1)

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Solution

The correct option is D $n (n + 1)$
Sum of first $n$ natural numbers $1, 2, 3, . . . . . ., n$
$S_{n} = 1 + 2 + 3 + . . . . . . . . . + (n - 1) + n . . . (i) S_{n} = n + (n - 1) + . . . . . . . . . . . . . + 2 + 1 . . . (i i)_________________________________{2 S}_{n} = (n + 1) + (n + 1) + . . . . . . + (n + 1) + (n + 1)$

Such $n$ terms
$2 S_{n} = n (n + 1) => S_{n} = \frac{n}{2} (n + 1)$
$S_{e v e n} = 2 + 4 + 6 + . . . . . + 2 (n - 1) + 2 n$
$S_{N} = 2 [1 + 2 + 3 + . . . . . + (n - 1) + n]$
$= \frac{2 n (n + 1)}{2}$
$= n (n + 1)$

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$1 + 3 + 5 + . . . . . + (2 n - 1) = n^{2}$ i.e., the sum of first n odd natural numbers is $n^{2}$ .

1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + 5^{2} + {2.6}^{2} + \dots . i s \frac{n}{2} (n + 1)^{2}

1.3.5 . . . . . (2 n - 1) > n^{n}, n ϵ N

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