1+3+5+...+(2n-1)?

Question

and i have to find each sum..i already know the answers, all i need to know is how do i work this...:D

Answer 1

assume S = sum of the numbers
assume n = n number
assume a = first number
assume d = the difference between the next and before number

first number is (a)
second number is (a + d)
third number is (a + 2d)
fourth number is (a + 3d)
n number is [a + (n - 1)d]

(a), (a + d), (a + 2d), (a + 3d), ..................[a + (n - 1)d]

the sum of n number is

equation (1)
S = (a) + (a + d) + (a + 2d) + (a + 3d) + ...........+ [a + (n - 1)d]

equation (2) sum n of number from the last number
S = [a + (n - 1)d] +............+ (a + 3d) + (a + 2d) + (a + d) + (a)

equation (1) + equation (2)
2S = {[2a + (n - 1)d] + [2a + (n - 1)d] +..............+ [2a + (n - 1)d]}

[a + (n - 1)d] repeat n times to get 2S

so, 2S = n[2a + (n - 1)d]
S = (n/2)[2a + (n - 1)d]

from the information that u you gave, a = 1 and d = 2
substitute a = 1 and d = 2 to S = (n/2)[2a + (n - 1)d]

so, S = (n/2)[2(1) + (n - 1)(2)]
= (n/2)[2 + 2n - 2]
= (n/2)(2n)
= n^2

the answer is 1+3+5+...+(2n-1) = n^2

you can try whether that is a equation by substitude the number of n.

for example, the sum for the first 3 numbers is 1 + 3 + 5 = 9
by using the equation you can get n^2 = 3^2 = 9

the sum for the first 4 numbers is 1 + 3 + 5 + 7= 16
by using the equation you can get n^2 = 4^2 = 16

the sum for the first 5 numbers is 1 + 3 + 5 + 7 + 9= 25
by using the equation you can get n^2 = 5^2 = 25

from the example above, we can notice that the equation S = n^2 is correct.

i hope you will understand what i have explained.

Answer 2

As other have said, you can add the end pair together to get 2n and there are (1/2)*n pairs for a sum of n*n.

But you can also do this visually.
Draw a square. Now add to the square by drawing squares to the top and right of the square. You added 3 to get a 2*2 square.

Add more squares(5) to get a 3*3 square.

As you add the odds numbers, as you add the nth odd number, you get n*n.


And now for some extra fun.

If (2n-1) is an odd perfect square(such as 9 or 25)

1+3+5 +7 +9 =25
(1+3+5+7) +9 =25
16 +9 = 25 This is a Pythagorean triplet

1+3 +5+7+9+11+13+15+17+19 +21+23+25 =169
(1+3 +5+7+9+11+13+15+17+19 +21+23)+25 =169
144 + 25 =169 This is also a Pythagorean triplet

This is a proof that there are an infinite number of Pythagorean triplets

Answer 3

[1+3+...+(2n-3)+(2n-1)]
+
[(2n-1)+(2n-3)+..+3+1]

If we add up pairs of terms like that, we end up with 2n + 2n + .. + 2n, repeated n times.
So double the sum is 2n*n, so the sum is n*n.

Answer 4

if u already know the answers why are u asking us but here's how to do it in order

P arentheses
E xponents
M ultiplication
D ivision
A ddition
S ubtraction

so when working an algebra problem just remember PEMDAS!

Answer 5

Use induction.
n, n, Sn
---------
1,1
2, 3, 4
3, 5, 13
4, 7, 20

try to find a pattern, then use induction...

Answer 6

sum is the answer :? and its 9+what+(2xn-1)