how do I prove the following: The sum of the first n odd positive integers = n^2; that is 1+3+...+(2n-1)=n^2?

Answer 1

Ha...lots of nice suggestions...but no work.

OK.

Basis case, n = 1:

1 = 1², check.

Induction:

Assume 1 + 3 + ... (2k - 1) = k².

Then

1 + 3 + ... + (2k - 1) + (2(k + 1) - 1) =

(using inductive hypothesis)

k² + (2k + 2 - 1) =

k² + 2k + 1 =

(k + 1)²

Which is what we wanted to show, so it's proved.

Answer 2

As the others have said, this is proved by mathematical induction. In fact, you can't ask for a more straightforward induction proof:

(1) Test the first case. OK, 1 = 1. Phase One complete.

(2) Assuming the nth case is true, prove it is true for the (n+1) case:

Assuming 1 + 3 + ... + (2n-1) = n^2, then 1 + 3 + ... + (2n-1) + (2n+1) = n^2 + (2n+1) = (n + 1)^2

...and you're done.

Answer 3

Sum of terms of arithmetic progression
= (first term + last term) / 2 x number of terms
= [ (1) + (2n - 1)] / 2 x n
= [1 + 2n - 1] / 2 x n
= [2n] / 2 x n
= n x n
= n^2

Alternatively
Sum
= [2 x first term + (number of terms - 1) x difference] x number of terms / 2
= [2 x 1 + (n - 1) x 2] x n /2
= [2 + 2n - 2] x n / 2
= [2n] x n / 2
= n x n
= n^2

Answer 4

n (n²-a million) = n*(n - a million)*(n + a million) n = 2*m + a million for n to be an difficulty-free integer, the position m is an integer. (2*m + a million)*(2*m*)*2*(m + a million) 4*(2*m + a million)*m*(m + a million) If that is divisible by technique of 24, then (2*m + a million)*m*(m + a million) should be divisible by technique of 6 (A) if m = 3*p, (2*m + a million)*m*(m + a million) = (6*p + a million)*(3*p)*(3*p + a million) If p is difficulty-free, the merely correct time period is even, and the expression is divisible by technique of two and by technique of three If p is even, the middle time period is even, and the expression is divisible by technique of two and by technique of three (B) if m = (3*p + a million), (2*m + a million)*m*(m + a million) = (6*p + 3)*(3*p + a million)*(3*p + 2) = 3*(p + a million)*(3*p + a million)*(3*p + 2) If p is difficulty-free, the first time period is even, and the expression is divisible by technique of two and by technique of three If p is even, the merely correct time period is even, and the expression is divisible by technique of two and by technique of three (C) if m = (3*p + 2), (2*m + a million)*m*(m + a million) = (6*p + 5)*(3*p + 2)*(3*p + 3) = 3* (6*p + 5)*(3*p + 2)*(p + a million) If p is difficulty-free, the merely correct time period is even, and the expression is divisible by technique of two and by technique of three f p is even, the middle time period is even, and the expression is divisible by technique of two and by technique of three if m = 3*p + 3..that's lower back a numerous of three, and case (A) applies lower back. QED.

Answer 5

It is trivial to prove using mathematical induction

Answer 6

one way is by mathematical induction

Answer 7

Do it by induction.