Question:

Let $a_n$ be the $n$th term of $5+8+14+23+35+50+\ldots$ and $S_n=\sum_{k=1}^n a_k$. Then $S_{30}-a_{40}$ is equal to

11310

11260

11290

11280

Step-by-Step Solution

Key Concept: $a_n=\frac{3n^2-3n+10}{2}$ by differences. $a_{40}=\frac{3(1600)-120+10}{2}=2345$. $S_{30}=13635$.

$11290$.

Correct Answer: 3