The sum $3 + 8 + 16 + 27 + 41 + ... $ upto 20 terms is equal to
Step-by-Step Solution
Key Concept: The sum of cubic and polynomial series can be decomposed using standard formulas for $\sum k^3$, $\sum k^2$, and $\sum k$
The terms are $t_n = n^3 + n^2 + n$. To find $S_n$, subtract the series from itself shifted: $S_n - S_{n-1}$ gives the $n$-th term. Computing systematically: $S_n = \sum_{k=1}^{n}(k^3 + k^2 + k) = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$. For $n = 20$: $S_{20} = \frac{20^2 \cdot 21^2}{4} + \frac{20 \cdot 21 \cdot 41}{6} + \frac{20 \cdot 21}{2} = 44100 + 2870 + 210 = 4430$.
Correct Answer: 4430