999 odd numbers

$\large \text {First lets generalize this for n using the following formula:}$

$\large \color{#D61F06} S_n= \frac{n}{2}[2a\ +\ (n - 1) d]$

$\large \text {In the arithmetic sum: 1 + 3 + 5 + 7 + ... + n; a = 1, d = 2, n =n (because we are generalizing this for n)}$

$\large \color{#20A900} S_n= \frac{n}{2}[2\ +\ (n - 1) 2] \\ = \large \color{#20A900}S_n= \frac{n}{2}[2\ +2n - 2] \\ \large \color{#20A900} =\ S_n= \frac{1}{2}n \times 2n \\ \large \color{#20A900} =\ S_n= \frac{2n^2}{2} \\ \large \color{#3D99F6} \boxed{= n^2}$

$\large \text {Since we have generalized it for n, we can now simply insert the data that we have into this equation}$

$\large \color{#EC7300} \therefore \boxed{999^2 = 998001}$

The sum of the first $n$ odd numbers is

$S(n)=\displaystyle\sum_{k=1}^n 2k-1$

To find a closed form

$\begin{aligned} S(n)&=\displaystyle\sum_{k=1}^n 2k-1\\ &=\displaystyle\sum_{k=1}^n 2k-\displaystyle\sum_{k=1}^n 1\\ &=2\displaystyle\sum_{k=1}^n k-\displaystyle\sum_{k=1}^n 1\\ &=2\cdot\dfrac{n(n+1)}{2}-n=n(n+1)-n=n^2 \end{aligned}$

In this case $n=999$

$S(999)=999^2=\boxed{998001}$

The $n$ th odd or even number is equal to $2n - 1$ , or $2n$ .

So, 999th odd is 2 $\times$ 999 - 1 = 1997.

The mean of 1 and 1997 is 999, so the answer is equal to $999 + 1998 \times 499$ .

$999 + 2 \times 999 \times 499 = 999 \times ( 1 + 2 \times 499 ) = 999 \times ( 1 + 998 ) = 999 \times 999 = 999 ^ { 2 } = ( 1000 - 1 ) ^ { 2 } = 1000 ^ { 2 } - 2 \times 1000 \times 1 + 1 ^ { 2 } = 1000000 - 2000 + 1 = 998001$

FORMULA --> square of n=998001