The question ask us to guess an explicit formula for the sequence
$$s_k = s_{k-1} + 2k ,$$ for all integers $k$ greater than or equal to one and $s_0 = 3$
Can someone help me with this? Because I don't really understand how to do this. Any help will be appreciated.
On
Hint: Moving the $s_{k-1}$ term to the LHS of the equation, the recurrence relation reads:
$$s_{k}-s_{k-1}=2k.$$
You can solve this recurrence relation by simply summing both sides over $k$:
$$\sum_{k=1}^{\infty}(s_{k}-s_{k-1})=2\sum_{k=1}^{\infty}k.$$
The sum on the LHS telescopes, while the sum on the RHS is a straightforward arithmetic progression. There are standard methods for finding nice closed forms for both series.
Rearrange to get:
$$s_k - s_{k - 1} = 2k$$
If we do a summation on both sides, a humongous amount of cancellation occurs on the LHS:
$$\sum_{i = 1}^{k} (s_i - s_{i - 1}) = \sum_{i = 1}^k 2i$$ $$s_k - s_0 = 2\sum_{i = 1}^ki$$
The RHS is easy to evaluate. Now just shift $s_0$ over to get $s_k$.