Question

The sum 1 + 3 + 11 + 25 + 45 + 71 + ... upto 20 terms, is equal to

• A. 7240
• B. 7130
• C. 6982
• D. 8124

Hint:

If the first differences of a series form an arithmetic progression (AP), then the general term of the series can be represented by a quadratic equation \( T_n = an^2 + bn + c \).

Answer 1

The Correct Option is A

To find the sum of the series \(1 + 3 + 11 + 25 + 45 + 71 + \ldots\) up to 20 terms, we start by analyzing the pattern in the given sequence.

Let's observe the differences between successive terms:

• 3 - 1 = 2
• 11 - 3 = 8
• 25 - 11 = 14
• 45 - 25 = 20
• 71 - 45 = 26

We can observe that the differences themselves form an arithmetic sequence: 2, 8, 14, 20, 26, etc., with a common difference of 6.

The general term of this sequence can be expressed as:

\(T_n = 1 + \sum_{i=1}^{n-1} \left(2 + (i-1) \times 6\right)\)

Simplify as follows:

\(T_n = 1 + \sum_{i=1}^{n-1} \left(6i - 4\right) = 1 + 6\sum_{i=1}^{n-1} i - 4(n-1)\)

Using the formula for the sum of the first \(n-1\) natural numbers, \(\sum_{i=1}^{n-1} i = \frac{(n-1) \cdot n}{2}\), we have:

\(T_n = 1 + 3(n-1)n - 4(n-1)\)

So,

\(T_n = 1 + (3n^2 - 3n - 4n + 4) = 3n^2 - 7n + 5\)

This is the nth term formula for the given sequence. Now, we need to find the sum of the first 20 terms:

\(S = \sum_{n=1}^{20} T_n = \sum_{n=1}^{20} (3n^2 - 7n + 5)\)

Breaking it down:

\(S = 3\sum_{n=1}^{20} n^2 - 7\sum_{n=1}^{20} n + \sum_{n=1}^{20} 5\)

Using the formulae:

• \(\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}\)
• \(\sum_{n=1}^{N} n = \frac{N(N+1)}{2}\)
• \(\sum_{n=1}^{N} 5 = 5N\)

Substituting these expressions for \(N = 20\):

• \(\sum_{n=1}^{20} n^2 = \frac{20 \times 21 \times 41}{6} = 2870\)
• \(\sum_{n=1}^{20} n = \frac{20 \times 21}{2} = 210\)
• \(\sum_{n=1}^{20} 5 = 5 \times 20 = 100\)

Thus, the sum \(S\) is:

\(S = 3 \times 2870 - 7 \times 210 + 100 = 8610 - 1470 + 100 = 7240\)

Therefore, the sum of the series up to 20 terms is 7240.

Answer 2

Given sum is \( S_n = 1 + 3 + 11 + 25 + 45 + 71 + ... + T_n \) 
First order differences are in A.P. 
Thus, we can assume that \( T_n = an^2 + bn + c \) 
Solving \( \begin{cases} T_1 = 1 = a + b + c T_2 = 3 = 4a + 2b + c \\T_3 = 11 = 9a + 3b + c \end{cases} \) 
we get a = 3, b = -7, c = 5 
Hence, general term of given series is \( T_n = 3n^2 - 7n + 5 \) 
Hence, required sum equals \( \sum_{n=1}^{20} (3n^2 - 7n + 5) = 3\frac{20 \cdot 21 \cdot 41}{6} - 7\frac{20 \cdot 21}{2} + 5(20) = 7240 \)