Question

Using the given pattern, find the missing numbers.

$$\begin{gathered} 1^2+2^2+2^2=3^2 \\ {2}^2+3^2+6^2=7^2 \\ {3}^2+4^2+{12}^2={13}^2 \\ {4}^2+5^2+{\_}^2={21}^2 \\ {5}^2+{\_}^2+{30}^2={31}^2 \\ {6}^2+7^2+{\_}^2={\_}^2 \end{gathered}$$

Answer 1

Method 1:

Let's find the missing squares in the pattern

The third number is the product of the first two numbers and the fourth number is obtained by adding 1 to the third number

$$\begin{gathered} 1^2+2^2+2^2=3^2 \\ {2}^2+3^2+6^2=7^2 \\ {3}^2+4^2+{12}^2={13}^2 \\ {4}^2+5^2+{20}^2={21}^2 \\ {5}^2+6^2+{30}^2={31}^2 \\ {6}^2+7^2+{42}^2={43}^2 \end{gathered}$$

For the given pattern, the missing squares in the pattern are 20, 6 and 42 respectively.

Alternative Method :

Here in this pattern, the third number which is squared in the pattern = product of the first two numbers which are squared.
The relation between the third number and the right-hand side number is that the right-hand side number is 1 more than the third number.
For example:
$5^2+6^2+{30}^2={31}^2$
In this case $5\times 6=30\mathrm{and}30+1=31$
Therefore, The missing numbers are as follows

:$$\begin{gathered} 4^2+5^2+{20}^2={21}^2 \\ {5}^2+6^2+{30}^2={31}^2 \\ {6}^2+7^2+{42}^2={43}^2 \end{gathered}$$
For the given pattern, the missing squares in the pattern are 20, 6 and 42 respectively.