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प्रश्न
For every natural number m, (2m – 1, 2m2 – 2m, 2m2 – 2m + 1) is a pythagorean triplet.
विकल्प
True
False
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उत्तर
This statement is False.
Explanation:
∵ (2m – 1)2 ≠ (2m2 – 2m)2 + (2m2 – 2m + 1)2
(2m2 – 2m)2 ≠ (2m – 1)2 + (2m2 – 2m + 1)2
And (2m2 – 2m + 1)2 ≠ (2m – 1)2 + (2m2 – 2m)2
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संबंधित प्रश्न
What will be the units digit of the square of the following number?
55555
What will be the units digit of the square of the following number?
53924
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?
(14, 48, 51)
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
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 :
52 + 62 + 72 + 82 + 92 + 102 + 112 + 122
Find the square of the following number:
451
Find a Pythagorean triplet in which one member is 12.
There are ______ natural numbers between n2 and (n + 1)2
The sides of a right triangle whose hypotenuse is 17 cm are ______ and ______.
