False. 15 x 15=225 and 15 x 15 x 15=3375. so the statement $\displaystyle { a }^{ 2 }$ ends in 5, then $\displaystyle { a }^{ 3 }$ ends in 25 is not true in all cases.

Let us take $\displaystyle x=1\times2\times3\times4=24 $

The squares of $431\;and\;7779$ would be odd numbers.
(Because we know that squares of odd numbers are always odd). 

The sum of the first odd numbers is,

$11 \times 11= 121$ 

5² = 5  × 5 = 25 = 12 + 13

$3^2$ means taking square of $3$, i.e. $3\times 3=9$

24 + 25 = 49 = 7²

361 = 19 × 19 = 19²
361 = 360 + 1 = 180 + 181

220 + 221 = 441 = 21  × 21 = 21²

Solving

Given Exp.$=\cfrac { { \left( a+b \right)  }^{ 2 }+{ \left( a-b \right)  }^{ 2 } }{ \left( { a }^{ 2 }+{ b }^{ 2 } \right)  } =\cfrac { 2\left( { a }^{ 2 }+{ b }^{ 2 } \right)  }{ \left( { a }^{ 2 }+{ b }^{ 2 } \right)  } =2$

$21600 $ can be factorized as $6\times 6\times 6\times 10\times 10$

According to Cauchy-Schwarz inequality:

First calculate the square-root of $594$

If a rectangle has perimeter 10 then the sum of its length and width is 5, giving two choice with integer sides:

$1^2+3^2+5^2+....+25^2$

The square of a number with 'n' number of zeroes, will have $ 2 \times n $ zeroes. Hence, there cannot be $ 5 $ zeroes in a square of a number as it would be an even number.

If the square of a number has $ 10 $ zeroes, the number will have $ 10 \div 2 = 5 $ zeroes.

If a number ends with $ 3 $ zeroes, then its square will have $ 3 \times 2 = 6 $ zeroes.

$49=7^2=$ sum of first $7$ odd numbers.
So, $49=1+3+5+7+9+11+13$.
Similarly, $121=11^2=$ sum of first $11$ odd numbers. 
So, $121=1+3+5+7+9+11+13+15+19+21$

$11^{2}$$=$$121$

$\sum _1^n(2n+1)=n^2$

Sum of odd numbers is $n^2$ where n is no. of odd nos.
n= 7 odd nos. 
Therefore sum is $7^2  = 49$

$13\times15$ $=$ $(10+3)$ $(10+5)$

Given series is $1+3+5+7+9+11+13+15+17+21$

Sum of odd natural numbers is $=n^2$

where $n= $ number of odd numbers are $11$

So, sum of the series given is $=11^2= 121 $.

$28 \times  28 = 78\underline{4} $
$162 \times  162 = 262\underline{44}$
(last digit is 4 even , so whole number is even.)

Therefore, A is the correct answer.

$7\times9$ $=$ $(10-3)$ $(10-1)$

$11\times13$ $=$ $(10+1)$ $(10+3)$

$44\times46$ $=$ $(40+4)$ $(40+6)$

Sum of first $8$ odd numbers is $1+3+5+7+9+11+13+15=64$

$N=$ Number of odd numbers are $8$, so $8^2 = 64$

Therefore, B is the correct answer.

$29\times31$ $=$ $(30-1)$ $(30+1)$

$16\times18$ $=$ $(20-4)$ $(20-2)$

Triangular numbers are obtained by continued summation of natural numbers $1,2,3,4,5,6,7,.....$ 

$22\times 24=(20+2)$$\times$$(20+4)$ 

$19\times 21=(20-1)$ $\times$ $(20+1)$ 

$14\times 16=(10+4)$$\times$$(10+6)$ 

$185$ $=$ $(200-15)$

$31\times 33=(30+1)$$\times$$(30+3)$ 

$11^{2}$ $=$ $(10+1)^{2}$ $=$ $100+1+20$

$15^{2}$ $=$ $(10+5)^{2}$ $=$ $100+25+100$

There are 4 consecutive odd numbers.
We know the formula for consecutive odd numbers of their sum = $n^2$
So, Sum = $4^2$ = 16
So, 1 + 3 + 5 + 7 = 16.

We know the formula for consecutive odd numbers of their sum $= n^2$
So, Sum of $13$ consecutive odd numbers $= 13^2 = 169$

Sum of first $100$ consecutive odd numbers $= 1 + 3 + 5 + 7 + 9 + 11 + ....+100 = 10000$
Since, we know the formula for sum of consecutive odd numbers $= n^2$
So, $n = 100$, Sum $= 100^2 = 10000$

Let the two consecutive numbers be, $\dfrac{n^{2} - 1}{2}$ and $\dfrac{n^{2} + 1}{2}$
$13^2= 169$
n = 13
$\dfrac{n^{2} - 1}{2} = \dfrac{13^{2} - 1}{2} = 84$
$\dfrac{n^{2} + 1}{2} = \dfrac{13^{2} + 1}{2} = 85$
So, the sum of two consecutive numbers = 84 + 85 = 169.

$21^{2}-1 = 400$
we can express the above number into general form $a^{2}-1 = (a + 1)\times (a - 1)$
Where a = 21
So, $21^{2}-1 = (21 + 1)\times (21 - 1)$
= $22 \times 20 = 400$
Therefore, the two even consecutive numbers are 20 and 22.

$11^{2}-1 = 120$
we can express the above number into general form $a^{2}-1 = (a + 1)\times (a - 1)$
Where a = 11
So, $11^{2}-1 = (11 + 1)\times (11 - 1)$
= $12 \times 10 = 120$
Therefore, the two even consecutive numbers are 10 and 12.

Let the two consecutive numbers be, $\dfrac{n^{2} - 1}{2}$ and $\dfrac{n^{2} + 1}{2}$
$15^2= 225$
n = 15
$\dfrac{n^{2} - 1}{2} = \dfrac{15^{2} - 1}{2} = 112$
$\dfrac{n^{2} + 1}{2} = \dfrac{15^{2} + 1}{2} = 113$
So, the sum of two consecutive numbers = 112 + 113 = 225.

$125^2 = 7813 + ?$
$15625 - 7813 = 7812$
So, the other consecutive number is $7812.$

$96^{2}-1 = 9215$
we can express the above number into general form $a^{2}-1 = (a + 1)\times (a - 1)$
Where a = 96
So, $96^{2}-1 = (96 + 1)\times (96 - 1)$
= $97 \times 95 = 9215$
Therefore, the two odd consecutive numbers are 95 and 97.

Sum of first 10 consecutive odd numbers $= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100$
Since, we know the formula for sum of consecutive odd numbers = $n^2$
So, $n = 10$, Sum $= 10^2 = 100$

$25^2 = 313 + ?$
$625 - 313 = 312$
So, the other consecutive number is $312$.

$12^2 = 144$
$102^2 = 10404$
$1002^2 = 1004004$
$10000002^2 = 100000040000004$
Start with 1 followed as many zeroes as there are between the first and the last 4, followed by two again followed by as many zeroes and end with 4.

From the pattern, the third number is the difference of the first two numbers.
The fourth number can be obtained by addition of the first two numbers.
Then, the missing numbers will be
$14^2 + 7^2 + 7^2 = 21^2$
So, $7^2, 21^2$ are the missing numbers.

From the pattern, the third number is the product of the first two number.
The fourth number can be obtained by adding 1 to the third number.
Then, the missing number will be
$4^2 + 3^2 + 12^2 = 13^2$
So, $13^2$ is the missing number.

From the pattern, the third number is the sum of the first two number.
The fourth number can be obtained by squaring the third number.
Then, the missing number will be
$5^2 + 2^2 +7^2 = 49^2$
So, $7^2$ is the missing number.

From the pattern, the third number is the division of the first two number.
The fourth number can be obtained by same as the first number.
Then, the missing numbers will be
$14^2 + 7^2 + 2^2 = 14^2$
So, $2^2, 14^2$ are the missing numbers.

$1^{2}$$=$$1$

$17^{2}$$-$$13^{2}$$=$$289-169$$=$$120$

Square numbers can only have EVEN NUMBER OF ZEROS because zero comes in only the square of $10$ and square of $10$ is $100$ that contains even number of zeros.

no of digits in a perfect square is $n$