Using the given pattern, find the missing numbers:
[2 marks]
$1^{2} + 2^{2} + 2^{2} = 3^{2}$
$2^{2} + 3^{2} + 6^{2} = 7^{2}$
$3^{2} + 4^{2} + 12^{2} = 13^{2}$
$4^{2} + 5^{2} + {- --- -}^{2} = 21^{2}$
$5^{2} + {- --- -}^{2} + 30^{2} = 31^{2}$
$6^{2} + 7^{2} + {- --- -}^{2} = {- --- -}^{2}$

Open in App

Solution

In each equation, the third number is the product of the first two numbers.
And the fourth number is obtained by adding 1 to the third number.
So,
$4^{2} + 5^{2} + 20^{2} = 21^{2}$ [0.5 mark]
$5^{2} + 6^{2} + 30^{2} = 31^{2}$ [0.5 mark]
$6^{2} + 7^{2} + 42^{2} = 43^{2}$ [1 mark]

Suggest Corrections

Similar questions

Question 6
Using the given pattern, find the missing numbers:
$1^{2} + 2^{2} + 2^{2} = 3^{2}$
$2^{2} + 3^{2} + 6^{2} = 7^{2}$
$3^{2} + 4^{2} + 12^{2} = 13^{2}$
$4^{2} + 5^{2} + {- --- -}^{2} = 21^{2}$
$5^{2} + {- --- -}^{2} + 30^{2} = 31^{2}$
$6^{2} + 7^{2} + - --- - 2 = - --- - 2$

Using the given pattern, find the missing numbers.
$1^{2} + 2^{2} + 2^{2} = 3^{2}$
$2^{2} + 3^{2} + 6^{2} = 7^{2}$
$3^{2} + 4^{2} + 12^{2} = 13^{2}$
$4^{2} + 5^{2} + . . .^{2} = 21^{2}$
$5^{2} + . . .^{2} + 30^{2} = 31^{2}$
$6^{2} + 7^{2} + . . . .^{2} = . . . .^{2}$ .

Q. Observe the following pattern

1^{2} = \frac{1}{6} [1 \times (1 + 1) \times (2 \times 1 + 1)] 1^{2} + 2^{2} = \frac{1}{6} [2 \times (2 + 1) \times (2 \times 2 + 1)] 1^{2} + 2^{2} + 3^{2} = \frac{1}{6} [3 \times (3 + 1) \times (2 \times 3 + 1)] 1^{2} + 2^{2} + 3^{2} + 4^{2} = \frac{1}{6} [4 \times (4 + 1) \times (2 \times 4 + 1)]

and find the values of each of the following:
(i) 1² + 2² + 3² + 4²+ ... + 10²
(ii) 5² + 6² + 7² + 8² + 9² + 10² + 11² + 12²

Q. Show by the Principle of Mathematical induction that the sum S_n of then terms of the series

1^{2} + 2 \times 2^{2} + 3^{2} + 2 \times 4^{2} + 5^{2} + 2 \times 6^{2} + 7^{2} + . . .

is given by

S_{n} = \{\begin{cases} \frac{n {(n + 1)}^{2}}{2}, if n is even \\ \frac{n^{2} (n + 1)}{2}, if n is odd \end{cases}

[NCERT EXEMPLAR]

Show by the Principle of Mathematical induction that the sum $S_{n}$ of the n terms of the series $1^{2} + 2 \times 2^{2} + 3^{2} + 2 \times 4^{2} + 5^{2} + 2 \times 6^{2} + 7^{2} + . . . . .$ is given by $S_{n} = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{n (n + 1)^{2}}{2}, if n is even \frac{n^{2} (n + 1)}{2}, if n is odd \end{matrix}$

Join BYJU'S Learning Program

Using the given pattern, find the missing numbers: [2 marks] 12+22+22=32 22+32+62=72 32+42+122=132 42+52+–––––2=212 52+–––––2+302=312 62+72+–––––2=–––––2

In each equation, the third number is the product of the first two numbers. And the fourth number is obtained by adding 1 to the third number. So, 42+52+202=212 [0.5 mark] 52+62+302=312 ​​​​​​​[0.5 mark] 62+72+422=432 ​​​[1 mark]

Using the given pattern, find the missing numbers:
[2 marks]
$1^{2} + 2^{2} + 2^{2} = 3^{2}$
$2^{2} + 3^{2} + 6^{2} = 7^{2}$
$3^{2} + 4^{2} + 12^{2} = 13^{2}$
$4^{2} + 5^{2} + {- --- -}^{2} = 21^{2}$
$5^{2} + {- --- -}^{2} + 30^{2} = 31^{2}$
$6^{2} + 7^{2} + {- --- -}^{2} = {- --- -}^{2}$

In each equation, the third number is the product of the first two numbers.
And the fourth number is obtained by adding 1 to the third number.
So,
$4^{2} + 5^{2} + 20^{2} = 21^{2}$ [0.5 mark]
$5^{2} + 6^{2} + 30^{2} = 31^{2}$ [0.5 mark]
$6^{2} + 7^{2} + 42^{2} = 43^{2}$ [1 mark]