Question

The math question is on the imagefind the nth term of the sequence ​

Answer 1

Notice how Pattern 2 is Pattern 1 with 4 balls added in the bottom row.

Pattern 3 is Pattern 2 with 5 more balls.

Pattern 4 is Pattern 3 with 6 more balls.

Generalizing the trend, we expect Pattern $n$ to be identical to Pattern $n-1$ with $n+2$ more balls.

If $b_n$ is the number of balls in the $n$-th pattern, then we have the recursive relation

$\begin{cases} b_1 = 6 \\ b_n = b_{n-1} + n + 2 & \text{for } n>1 \end{cases}$

We can solve this recurrence by substitution. Using the definition of $b_n$, we have

$b_{n-1} = b_{n-2} + (n-1) + 2 \\\\ \implies b_n = (b_{n-2} + (n-1) + 2) + n + 2 \\\\ \implies b_n = b_{n-2} + 2\times2 + \bigg(n + (n-1)\bigg)$

$b_{n-2} = b_{n-3} + (n-2) + 2 \\\\ \implies b_n = (b_{n-3} + (n-2) + 2) + 2\times 2 + \bigg(n + (n-1)\bigg) \\\\ \implies b_n = b_{n-3} + 3\times2 + \bigg(n + (n-1) + (n-2)\bigg)$

and so on, down to

$b_n = b_1 + (n-1)\times2 + \bigg(n + (n-1) + (n-2) + \cdots + 2\bigg)$

Recall that

$\displaystyle \sum_{i=1}^n i = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

Then we find

$\displaystyle b_n = 6 + 2(n-1) + \sum_{i=2}^n i$

$\displaystyle b_n = 2n + 4 + \left(\sum_{i=1}^n i - 1\right)$

$\displaystyle b_n = 2n + 3 + \frac{n(n+1)}2$

$\displaystyle \boxed{b_n = \frac{n^2+5n+6}2}$