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प्रश्न
Write the Pythagorean triplet whose one of the numbers is 4.
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उत्तर
For any natural number greater than 1, (2m, m2 – 1, m2 + 1) is Pythagorean triplets.
So, if one number is 2m, then another two numbers will be m2 – 1 and m2 + 1.
Given, one number = 4
Then Pythagorean triplets:
2m = 4 or m = 2
So, m2 – 1 = (2)2 – 1 = 4 – 1 = 3
m2 + 1 = (2)2 + 1 = 4 + 1 = 5
Now, (3)2 + (4)2 = (5)2
⇒ 9 + 16 = 25
⇒ 25 = 25
Therefore, Pythagorean triplets are 3, 4 and 5.
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संबंधित प्रश्न
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\[\left( 1 \times 2 \right) + \left( 2 \times 3 \right) = \frac{2 \times 3 \times 4}{3}\]
\[\left( 1 \times 2 \right) + \left( 2 \times 3 \right) + \left( 3 \times 4 \right) = \frac{3 \times 4 \times 5}{3}\]
\[\left( 1 \times 2 \right) + \left( 2 \times 3 \right) + \left( 3 \times 4 \right) + \left( 4 \times 5 \right) = \frac{4 \times 5 \times 6}{3}\]
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Observe the following pattern \[1 = \frac{1}{2}\left\{ 1 \times \left( 1 + 1 \right) \right\}\]
\[ 1 + 2 = \frac{1}{2}\left\{ 2 \times \left( 2 + 1 \right) \right\}\]
\[ 1 + 2 + 3 = \frac{1}{2}\left\{ 3 \times \left( 3 + 1 \right) \right\}\]
\[1 + 2 + 3 + 4 = \frac{1}{2}\left\{ 4 \times \left( 4 + 1 \right) \right\}\]
and find the values of following:
1 + 2 + 3 + 4 + 5 + ... + 50
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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