Question

The sum of the first 20 terms of the series 5 + 11 + 19 + 29 + 41 + ... is

Answer 1

Correct Answer: 3520

The answer is \(3520\)

\(T_{n}=an^{2}+bn+c\)
\(T_{1}=a+b+c=5\) 
\(T_{2}=4a+2b+c=11\)
\(T_{3}=9a+3b+c=19\)
\(T_{n}=n^{2}+3n+1\)
\(S_{20}=\sum_{n=1}^{20}T_{n}=\sum_{n=1}^{20}(n^{2}+3n+1)\)
\(S_{20}=\frac{20\times 21\times 41}{6}+3\times\frac{20\times 21}{2}+20\)
\(S_{20}=2870+630+20 \)
\(S_{20}=3520\)

Answer 2

Series Sum Calculation 

Let the given series be denoted by $S_n = 5 + 11 + 19 + 29 + 41 + \cdots + T_n$, where $T_n$ is the $n^{th}$ term. Let's find the differences between consecutive terms:

$11 - 5 = 6$

$19 - 11 = 8$

$29 - 19 = 10$

$41 - 29 = 12$

The differences form an arithmetic progression with the first term $a = 6$ and common difference $d = 2$.

The $n^{th}$ term of this arithmetic progression is given by $a_n = a + (n - 1)d = 6 + (n - 1)2 = 6 + 2n - 2 = 2n + 4$.

The $n^{th}$ term of the original series is given by the sum of the terms in the arithmetic progression up to $n-1$ terms plus the first term 5.

$T_n = 5 + \sum_{k=1}^{n-1} (2k + 4) = 5 + 2\sum_{k=1}^{n-1} k + 4(n - 1)$

$T_n = 5 + 2 \frac{(n - 1)(n)}{2} + 4(n - 1) = 5 + n^2 - n + 4n - 4 = n^2 + 3n + 1$

We want to find $S_{20}$, the sum of the first 20 terms.

$S_{20} = \sum_{n=1}^{20} T_n = \sum_{n=1}^{20} (n^2 + 3n + 1) = \sum_{n=1}^{20} n^2 + 3\sum_{n=1}^{20} n + \sum_{n=1}^{20} 1$

Using the formulas for the sum of the first N integers and the sum of the first N squares:

$\sum_{n=1}^{N} n = \frac{N(N + 1)}{2}$

$\sum_{n=1}^{N} n^2 = \frac{N(N + 1)(2N + 1)}{6}$

$S_{20} = \frac{20(21)(41)}{6} + 3\frac{20(21)}{2} + 20 = 2870 + 630 + 20 = 3520$

Conclusion: The sum of the first 20 terms is 3520.