You may use a calculator. But how?

Here's a simple step to step procedure

$1.$ On your calculator $\text{(Most important!!)}$

$2.$ Press 2,then 0, then 1, and then 6

$3.$ Press ' $\times$ ' and then repeat $2.$ followed by pressing '='.

$4.$ Type that 7 magical digits in the answer's column and see the magic...Thanks

Using Arithmetic Progression

1,3,5,7.......(2n-1) where n=2016, a=1, d=2

We need to take out the sum of this AP, so

$S_{n}=\frac{n}{2}[ 2.a+(n-1).d ]$

$S_{2016}=\frac{2016}{2}[ 2.(1)+(2016-1).2 ]$

$S_{2016}=1008[ 2+4030]$

$S_{2016}=1008.4032$

$S_{2016}=4064256$

Thus, the sum will be 4064256 which is nothing but $(2016)^{2}$ .

$\rightarrow$ $\text{Sum of odd numbers}=$ $n^2$
$\Rightarrow 2016^2=\boxed{4064256}$

Solution:

I will solve this with the mathematical induction. When that number ( $2016$ ) is $1$ the answer is $1$ . When it is $2$ , the answer is $4 (1 + 3)$ , for $3$ it is $9 (1 + 3 + 5)$ and so on. We can see that for a number $n$ , answer should be $n^{2}$ , which is true. The following is just a "proof".

If we believe that for the first $n$ odd numbers sum is $n^{2}$ , we only need to prove that this will work for $n + 1$ In the latter case, sum is: $1 + 3 + 5 + ..... + 2n - 1 + 2\times(n + 1) - 1$ .

The first part: $1 + 3 + 5 + ..... + 2n - 1$ is the sum of the first $n$ numbers which is $n^{2}$ . Inserting that we get: $n^{2} + 2\times(n + 1) - 1 = n^{2} + 2n + 1 =$ $n\times n + n + n + 1 = n \times (n + 1) + n + 1 = (n + 1)(n + 1) = (n + 1)^{2}$ .

With this, the mathematical induction is complete and the solution will really be $2016^{2} = \boxed{4 064 256}$ .

by the formula of serious for every n digit odd numbur sum, gotta find the answer by n^2. For 2016 th odd number it's process 2016^2=4064256.

if we get first 2 odd numbers.they are 1,3 and their sum is 4=2^2, for first 3 odd numbers, sum is 9=3^2 so we can say sum of first 'n' odd numbers = n^2 So for 2016 it must be 2016^2 = 4064256

Its just that, Sum of first n odd no:s is n^2..... so here n=2016.

Using arithmetric progression whose a1 = 1 and r = 2.

a1, a2, a3,...., a2016

1, 3, 5,...., a2016

an = a1 + (n-1) • r

a2016 = 1 + (2016-1) • 2

a2016 = 4031

Sn = ((a1+an)÷2) • n

Sn = ((1 + 4031)÷2) • 2016

Sn = 2016 • 2016

Sn = 2016²

Sn = 4.064.256

1+3+5+...up to 2016 terms=(2016)^2=4064256

For adding odd numbers use n^2 where n=no.of odd numbers to be added

I forgot about n^2. I noticed that the sum of 1st and nth term is equal to 2n. The 2nd and n-1 terms do the same. Because n is even, the first half of the numbers all have a paired term that equals 2n.

So: (2n)(n/2)=n*n

OR. ... n^2 lol but again I forgot this fact.

Sum of n odd numbers equal to n^2

If you didn't remember that the sum of the first n , odd integers is $n^2$ ,

First note that the sum of the first n odd integers will be n less than the sum of the first n even integers.

Included in the sum of the first 4032 integers is the sum of the first 2016 even and odd integers.

The sum of the first n integers is n(n+1)/2 , so

$4032(4032+1)/2=S_{even} +S_{odd}$

$S_{even} =S_{odd} +2016$ ,

$4032(4032+1)/2=S_{odd} +2016+S_{odd}$

$2016(4032+1)=2S_{odd}+2016$

$2016(2016*2)+2016=2S_{odd}+2016$

$S_{odd} =2016^2=4064256$

Sum of an AP = n/2(a+l) (where 'a' is first and 'l' is last term)

We know that last term = a + (n-1)d -------> (i)

Therefore,

Sum of AP = n/2[a + a + (n-1)d] (Using (i))

               = n/2[2a + (n-1)d]   -------> (ii)

Here,

No of terms(n) = 2016

First Term(a) = 1

Common Difference(d) = 2 (because between any two odd numbers there is an even number)

Using these in (ii),

2016/2[(2x1) + (2016-1)x2 ]

2016/2[2 + (2016-1)x2 ]

Taking 2 as common,

2016/2 x 2[1 + (2016-1)]

2016/2 x 2[1 - 1 + 2016]

2016 X 2016

=4064256

And for any questions relating to finding the sum of 'n' number of odd terms, use the formula (n)^2....... it works....

This is the way I was first taught how to solve these, and it's pretty easy to remember in my opinion:

Bear in mind that the formula for adding consecutive numbers, 1 to n is n (n+1)/2 So

1 = 1×2-1

3=2×2-1

5=3×2-1

Etc, and, following the pattern*, the last number must be Last number = 2016×2-1

Now to add them. First factorise by pulling a two out (as they've all got a ×2) and you get

2 (1+2+3+...+2016)-2016 (as there's 2016 -1s, due to there being 2016 numbers) Finally, by using the formula mentioned above we get that the sum equals

2 ×2016 (2016+1)/2-2016 The 2s cancel out so we have 2016×2017-2016 Which, if we punch into a calculator should give us the answer of

4064256

*notice how the reason I wrote the numbers like this is so that we'd have a set of consecutive numbers (the 1 2 3 etc) timesed by the same number throughout, then the same number taken away throughout so that we could add them all up, as factorised using the ×2 and simply added the 1s up by timsing by how many there are.