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प्रश्न
The sum of first n odd natural numbers is ______.
पर्याय
2n + 1
n2
n2 – 1
n2 + 1
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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
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संबंधित प्रश्न
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From the pattern, we can say that the sum of the first n positive odd numbers is equal to the square of the n-th positive number. Putting that into formula:
1 + 3 + 5 + 7 + ... n = n2, where the left hand side consists of n terms.
Which of the following triplet pythagorean?
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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
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