The nth term of a sequence is n^2+n

Try factor out the nth term formula, what do you notice?

The nth term of a sequence is n^2+n
Two consecutive terms in the sequence have a difference of 32.
Work out the two terms.

Thank you!!

n(n+1)??
I don't notice anything. 😱😱😱

Two consecutive terms are $n$ and $n+1$. We are looking for a pair so that the difference is 32. This means we can formulate $[(n+1)^2 +(n+1)] - [n^2 + n] = 32$

Aren't the contents of the first brackets multiplied instead of summed?

Two consecutive terms are $n$ and $n+1$. We are looking for a pair so that the difference is 32. This means we can formulate $[(n+1)^2 +(n+1)] - [n^2 + n] = 32$

Also, Im pretty sure it should be (n+1)(n+2)-n (n+1)

Why would they be multiplied?

It means that a term of a sequence is found by multiplying it by number right after it. Eg: the 7th term is found by multiplying 7 and 8. Can you now work out how to do it?

Please, another clue thank you. I don't get it...

Nvm ignore me, you're right.

The nth term of a sequence is n^2+n
Two consecutive terms in the sequence have a difference of 32.
Work out the two terms.

Thank you!!

let's say the first term if the two consecutive numbers is n. this means that's the next number is n+1.
now since the difference is 32,we can say;
nth term - (n+1)th term = 32
And we r givene the formula to calculate the nth term of the sequence so;
n (n+1) - (n+1)(n+1+1) = 32
n(n+1)-(n+1)(n+2) = 32.
Can u do it from here ?

Have you tried what I said?

Is it 15 and 16?

Yep.

You mean (n+1)th term minus nth term, not other way around