Mathematics: 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?

False Step-by-step explanation:Because any number below 0 is negative...

How to find nth term for a sequence of following numbers;

oliviasoreo92

10.03.2022

Let a_n denote the given sequence. a_n 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 b_n , then for $n\ge1$ ,

b_n=(n+1)^3

b_n is defined in terms of a_n for all $n\ge1$ by

$b_n=a_{n+1}-a_n$

and so a_n is defined recursively by

$a_n=\begin{cases}a_1=1\\a_{n+1}=a_n+(n+1)^3&\text{for }n\ge1\end{cases}$

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

a_2=a_1+2^3

$a_3=a_2+3^3=a_1+\displaystyle\sum_{i=1}^2(i+1)^3$

$a_4=a_3+4^3=a_1+\displaystyle\sum_{i=1}^3(i+1)^3$

and so on, up to

$a_n=a_1+\displaystyle\sum_{i=1}^{n-1}(i+1)^3$

We can simplify the right hand side a bit, noticing that a_1=1=1^3 matches (i+1)^3 for i=0 :

$a_n=\displaystyle\sum_{i=0}^{n-1}(i+1)^3$

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

$a_n=\displaystyle\sum_{i=1}^ni^3$

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

$a_n=\dfrac{n^2(n+1)^2}4$

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