How to find nth term for a sequence of following numbers; 1,9,36, i know that this is square sequence and i use the formula a+(n-1)d+1/2×(n-1)×(n-2)c where a=the first term of first sequence d=difference between first two terms in first sequence and c=common difference between terms in the second difference. how to express c in this example if the terms in the second sequence are cube numbers?

Let denote the given sequence. has forward differences

{9 - 1, 36 - 9, 100 - 36, ...} = {8, 27, 64, ...} = {2^3, 3^3, 4^3, ...}

If we call the sequence of forward differences , then for ,

is defined in terms of for all by

and so is defined recursively by

We can deduce a pattern for the general -th term:

and so on, up to

We can simplify the right hand side a bit, noticing that matches for :

and to simplify things a bit more, we shift the index of summation:

You should know that the right side has a nice closed form (look up "Faulhaber's formula" if you don't):

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step-by-step explanation:

22.83 repeating

step-by-step explanation:

137 ÷ 6 is 22.8333333 therefore we understand that 3 is repeating.

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