True or False?

Question

the result of multiplying the sum of the first n positive even integers by 4 and then adding 1 is always a perfect square.

Answer 1

True
look at this, if I take the sum of the first five positive even integers and do what you are saying I will get
4[x + x+2 + x+4 + x+6 + x+8] +1
4[5x + 20] + 1
20x + 80 + 1
20x + 81 now replace with 2 and you get
121 which is a perfect square

but if you take the first six positive event integers you will have
4[x + x+2 + x+4 + x+6 + x+8 + x+10] +1
4{6x + 30] + 1
24x + 120 + 1
24x + 121 and when you put x=2 you get 169
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Answer 2

Let's put it this way:

The sum of the first n positive even integers can be written with the following equation:

n(n+1) = n^2 + n

Simple enough. Next, multiply by 4:

4n^2 + 4n

And add 1:

4n^2 + 4n + 1

...which can be factored into (2n+1)^2.

So the result of multiplying the sum of the first n positive integers by 4 and then adding 1 is always a perfect square of base 2n+1, provided n is positive.

Answer 3

Well, the sum in question is twice the sum of the first n integers (just factor out a 2 from each term in the sum). I.e., it's 2 * n(n+1)/2 = n(n+1).

And 4n(n+1)+1 = (2n+1)^2, as has already been pointed out.

Answer 4

4 (2n + 2n+2 + 2n+4 + 2n+6....... ) +1
8n + 8n+8 +8n +16 + 8n +24 ..... +1

true