Advertisements
Advertisements
प्रश्न
The sum of first n odd natural numbers is ______.
पर्याय
2n + 1
n2
n2 – 1
n2 + 1
Advertisements
उत्तर
The sum of first n odd natural numbers is n2.
Explanation:
Sum of first n odd natural numbers
= `sum(2n - 1) = 2sumn - n`
= `(2 xx n(n + 1))/2 - n`
= n(n + 1) – n
= n2 + n – n
= n2
APPEARS IN
संबंधित प्रश्न
Write a Pythagorean triplet whose one member is 16.
Which of the following triplet pythagorean?
(16, 63, 65)
Observe the following pattern
\[\left( 1 \times 2 \right) + \left( 2 \times 3 \right) = \frac{2 \times 3 \times 4}{3}\]
\[\left( 1 \times 2 \right) + \left( 2 \times 3 \right) + \left( 3 \times 4 \right) = \frac{3 \times 4 \times 5}{3}\]
\[\left( 1 \times 2 \right) + \left( 2 \times 3 \right) + \left( 3 \times 4 \right) + \left( 4 \times 5 \right) = \frac{4 \times 5 \times 6}{3}\]
and find the value of(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + (5 × 6)
Observe the following pattern \[1^2 = \frac{1}{6}\left[ 1 \times \left( 1 + 1 \right) \times \left( 2 \times 1 + 1 \right) \right]\]
\[ 1^2 + 2^2 = \frac{1}{6}\left[ 2 \times \left( 2 + 1 \right) \times \left( 2 \times 2 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 = \frac{1}{6}\left[ 3 \times \left( 3 + 1 \right) \times \left( 2 \times 3 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 + 4^2 = \frac{1}{6}\left[ 4 \times \left( 4 + 1 \right) \times \left( 2 \times 4 + 1 \right) \right]\] and find the values :
12 + 22 + 32 + 42 + ... + 102
Which of the following number square of even number?
4489
Find the square of the following number:
862
Find the square of the following number:
425
If m is the square of a natural number n, then n is ______.
There are ______ natural numbers between n2 and (n + 1)2
