Question

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

Answer 1

Given $1^2+2^2+2^2=3^2$
$2^2+3^2+6^2=7^2$
$3^2+4^2+{12}^2={13}^2$
Here, the pattern we can identify is that the third number is the product of first and second number.
Because, for $1^2+2^2+2^2=3^2$
Here, the third number is $2=1\times 2$
The same rule holds for the next two equations as well.
Similar way, the fourth term is $1$ more than the third term.
Because, for $1^2+2^2+2^2=3^2$
Here, the fourth term is $3=2+1$
The same rule holds for the next two equation as well.
(i) Now for, $4^2+5^2+..{.}^2={21}^2$
Since, here third term is product of first terms .
That is third term is $=4\times 5=20$
Therefore, the given equation can be written as $4^2+5^2+{20}^2={21}^2$
(ii) Similar way for ,$5^2+..{.}^2+{30}^2={31}^2$
Since, the third term $(c=30)$ is product of first term ($a=5$) and second term ($b=?$).
Thus, $c=ab$
$\Rightarrow b=\frac{c}{a}$
$=\frac{30}{5}$
$\therefore b=6$
Hence, the given expression can be written as $5^2+6^2+{30}^2={31}^2$

(iii) For $6^2+7^2+...{.}^2=...{.}^2$
Third term is product of first two terms.
That is third term $=6\times 7=42$
The fourth term is one more than the third term.
That is fourth term is$=42+1=43$
Therefore, the given expression is $6^2+7^2+{42}^2={43}^2$