The sum of first n positive integers is given by

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

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

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

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None of these

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Solution

The correct option is A $\frac{n (n + 1)}{2}$
Standard formula.

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Similar questions

Q. If

n \geq 2

is a positive integer, then the sum of the series

^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + \dots +^{n} C_{2})

Find the sum of first $n$ natural numbers.

The sum of first n terms of the given series $1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + 5^{2} + {2.6}^{2} + . . . . . . . . . i s \frac{n (n + 1)^{2}}{2}$ , When n is even. When n is odd, the sum will be

Q. The sum of the series

1 \times n + 2 (n - 1) + 3 (n - 2) + \dots \dots + (n - 1) \times 2 + n \times 1

Q. If

n

is a positive integer, show that
(1)

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

;
(2)

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

;
the series in each case being extended to

n

terms; and
(3)

1^{n} - n 2^{n} + \frac{n (n - 1)}{1 \cdot 2} 3^{n} - \dots = {(- 1)}^{n} n!

;
(4)

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

;
the series in the last two cases being extended to

n + 1

terms.

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The sum of first n positive integers is given by

The correct option is A n(n+1)2Standard formula.

The correct option is A $\frac{n (n + 1)}{2}$
Standard formula.