Tag: some special sequences

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.

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