The sum of the series
`3+8+16+27+41`…………… upto 20 terms is equal to

Options

A. 4230
B. 4430
C. 4330
D. 4500

Answer

To find the sum of the series \(3 + 8 + 16 + 27 + 41 + \ldots\) up to 20 terms, we first need to identify the pattern in the series.

### Step 1: Identify the terms and their differences
The given series is:
- First term: \(a_1 = 3\)
- Second term: \(a_2 = 8\)
- Third term: \(a_3 = 16\)
- Fourth term: \(a_4 = 27\)
- Fifth term: \(a_5 = 41\)

Now, let's find the differences between consecutive terms:
- \(a_2 - a_1 = 8 - 3 = 5\)
- \(a_3 - a_2 = 16 - 8 = 8\)
- \(a_4 - a_3 = 27 - 16 = 11\)
- \(a_5 - a_4 = 41 - 27 = 14\)

The differences are \(5, 8, 11, 14\), which form an arithmetic progression (AP) with a common difference of \(3\).

### Step 2: Find the nth term of the series
The first difference can be expressed as:
- \(d_1 = 5\)
- \(d_2 = 8 = 5 + 3\)
- \(d_3 = 11 = 5 + 2 \cdot 3\)
- \(d_4 = 14 = 5 + 3 \cdot 3\)

The nth difference can be expressed as:
\[ d_n = 5 + (n-1) \cdot 3 = 3n + 2 \]

Now, we can express the nth term \(a_n\) of the series:
\[ a_n = a_1 + \sum_{k=1}^{n-1} d_k \]

Calculating the sum of the first \(n-1\) differences:
\[ a_n = 3 + \sum_{k=1}^{n-1} (3k + 2) \]
\[ = 3 + \left(3 \sum_{k=1}^{n-1} k + 2(n-1)\right) \]
Using the formula for the sum of the first \(m\) natural numbers, \(\sum_{k=1}^{m} k = \frac{m(m+1)}{2}\):
\[ \sum_{k=1}^{n-1} k = \frac{(n-1)n}{2} \]

Thus:
\[ a_n = 3 + 3 \cdot \frac{(n-1)n}{2} + 2(n-1) \]
\[ = 3 + \frac{3(n^2 - n)}{2} + 2n - 2 \]
\[ = 3 + \frac{3n^2 - 3n + 4n - 4}{2} \]
\[ = 3 + \frac{3n^2 + n - 4}{2} \]
\[ = \frac{6 + 3n^2 + n - 4}{2} \]
\[ = \frac{3n^2 + n + 2}{2} \]

### Step 3: Calculate the sum of the first 20 terms
Now, we can find the sum of the first \(N\) terms:
\[ S_N = \sum_{n=1}^{N} a_n \]
\[ S_N = \sum_{n=1}^{N} \frac{3n^2 + n + 2}{2} \]
\[ = \frac{1}{2} \left( 3 \sum_{n=1}^{N} n^2 + \sum_{n=1}^{N} n + 2N \right) \]

Using the formulas:
- \(\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}\)
- \(\sum_{n=1}^{N} n = \frac{N(N+1)}{2}\)

Substituting \(N = 20\):
\[ S_{20} = \frac{1}{2} \left( 3 \cdot \frac{20 \cdot 21 \cdot 41}{6} + \frac{20 \cdot 21}{2} + 40 \right) \]

Calculating each term:
1. \(3 \cdot \frac{20 \cdot 21 \cdot 41}{6} = 3 \cdot 2870 = 8610\)
2. \(\frac{20 \cdot 21}{2} = 210\)
3. \(40 = 40\)

Thus:
\[ S_{20} = \frac{1}{2} (8610 + 210 + 40) = \frac{1}{2} \cdot 8850 = 4425 \]

### Final Answer
The sum of the series up to 20 terms is \(4425\).