Tag: some special sequences

$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.