#1
4, 13, 26, 43....form a quadratic sequence. Prove that 859 is in this sequence.

I got the quadratic sequence as〖2n〗^2-n-1

But I don't know how to prove it
0
5 months ago
#2
It's actually 2n^2 + 3n - 1,

So for some value of n, 2n^2 + 3n - 1 = 859, what is this value of n??
0
5 months ago
#3
(Original post by Nasir.)
4, 13, 26, 43....form a quadratic sequence. Prove that 859 is in this sequence.

I got the quadratic sequence as〖2n〗^2-n-1

But I don't know how to prove it
If you have correctly obtained the formula for calculating x term in the sequence, just set your formula - 2n^2-n-1 according to you - equal to 859 and then rearrange to set it equal to zero. It should form an ordinary quadratic that you can solve, and you can go from there.
Last edited by Jackudy3; 5 months ago
0
5 months ago
#4
the sequence should be 2n^2+3n-1 to prove it you would make the equation equal to 859, minus 859 from both sides to make the quadratic equal to 0 then use the quadratic formula to show that n is an integer/whole number
0
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