Question

Using the given pattern, find the missing numbers in the pattern
\[
  {1^2} + {2^2} + {2^2} = {3^2} \\
  {2^2} + {3^2} + {6^2} = {7^2} \\
  {3^2} + {4^2} + {12^2} = {13^2} \\
  {4^2} + {5^2} + { \ldots ^2} = {7^2} \\
  {5^2} + { \ldots ^2} + {30^2} = {31^2} \\
  {6^2} + {7^2} + { \ldots ^2} = { \ldots ^2} \\
 \]

Answer 1

Hint: For analysis of this given pattern requires a mathematical and logical reasoning to observe the numbers occurring at various places.


Complete step-by-step answer:
Observing the first line of the given pattern. $\left( {{1^2} + {2^2} + {2^2} = {3^2}} \right)$
It is clear that the square of the first number, plus Whole Square of first number plus one, plus square of the multiplication of the first and second number gives a square of a number which is square of the number which is one plus the number obtained by multiplying first and second number.
Let’s suppose $n = 1$ , then the first line of the series becomes
First number as ${\left( n \right)^2} = {\left( 1 \right)^2}$
Second number as ${\left( {n + 1} \right)^2} = {2^2}$
Third number as ${\left[ {n\left( {n + 1} \right)} \right]^2} = {2^2}$
 The resultant number is obtained by adding all the three numbers is,
\[{\left[ {n\left( {n + 1} \right) + 1} \right]^2} = {\left( 3 \right)^2}\]
The given pattern in the form of n becomes
${n^2} + {\left( {n + 1} \right)^2} + {\left[ {n\left( {n + 1} \right)} \right]^2} = {\left[ {n\left( {n + 1} \right) + 1} \right]^2} \cdots \left( 1 \right)$
For the first line substitute $n = 1$ in equation (1), then it becomes
$
  {\left( 1 \right)^2} + {\left( {1 + 1} \right)^2} + {\left[ {1\left( {1 + 1} \right)} \right]^2} = {\left[ {1\left( {1 + 1} \right) + 1} \right]^2} \\
  {1^2} + {2^2} + {2^2} = {3^2} \\
 $
 For the second line substitute $n = 2$ in equation (1) , then it becomes
$
  {2^2} + {\left( {2 + 1} \right)^2} + {\left[ {2\left( {2 + 1} \right)} \right]^2} = {\left[ {2\left( {2 + 1} \right) + 1} \right]^2} \\
  {2^2} + {3^2} + {6^2} = {7^2} \\
 $
Similarly for third line of the put $n = 3$ in equation, it becomes
${3^2} + {4^2} + {12^2} = {13^2}$
For obtaining the missing term of fourth line substitute $n = 4$ in equation, it becomes as
${4^2} + {5^2} + {20^2} = {21^2}$
The missing term is$20$.

For obtaining the missing term of fifth line substitute $n = 5$ in equation, it becomes as
${5^2} + {6^2} + {30^2} = {31^2}$
The missing term is $6$.

For obtaining the two missing terms of sixth line substitute $n = 5$ in equation, it becomes as
${6^2} + {7^2} + {42^2} = {43^2}$
The missing term is $42$ and $43$.


Note: For observing any pattern, the critical observation of the terms of the series should be done.
Here in the question putting $n = 1$ in the first line satisfies the first line.
In the same way for successive lines substitution of the value of $\left( {n + 1} \right)$ will help to arrive at the whole pattern.