Question

If n^th term of a series is n+(-1)^n-1, n = 1,2,3,..., then the sum of first 40 terms of the series is

• A. 810
• B. 820
• C. 821
• D. 819
• E. 780

Answer 1

The Correct Option is B

The \( n \)-th term of the series is given by \( T_n = n + (-1)^{n-1} \).
We can split the series into two parts:
When \( n \) is odd, \( (-1)^{n-1} = 1 \), so \( T_n = n + 1 \).
When \( n \) is even, \( (-1)^{n-1} = -1 \), so \( T_n = n - 1 \).
Thus, the series alternates between terms of the form \( n+1 \) for odd \( n \) and \( n-1 \) for even \( n \).
We now need to find the sum of the first 40 terms. The sum of the terms for odd \( n \) is the sum of \( n+1 \), and for even \( n \), it is the sum of \( n-1 \).
For odd \( n \), the terms are: \( 1+1, 3+1, 5+1, \dots \), up to the 40th term. This gives us a sum of 820.

The correct option is (B) : \(820\)

Answer 2

The given nth term of the series is:

\[ T_n = n + (-1)^{n-1} \] where \(n = 1, 2, 3, \dots\).

Let's calculate the sum of the first 40 terms of the series:

We can break the expression for \(T_n\) into two parts:

\[ T_n = n + (-1)^{n-1} \] The sum of the first 40 terms of the series is: \[ S_{40} = \sum_{n=1}^{40} T_n = \sum_{n=1}^{40} \left(n + (-1)^{n-1}\right) \] This can be written as the sum of two separate series: \[ S_{40} = \sum_{n=1}^{40} n + \sum_{n=1}^{40} (-1)^{n-1} \] 1. The sum of the first 40 natural numbers: \[ \sum_{n=1}^{40} n = \frac{40(40+1)}{2} = 820 \] 2. The sum of the alternating series \(\sum_{n=1}^{40} (-1)^{n-1}\): - For even \(n\), \((-1)^{n-1} = -1\), - For odd \(n\), \((-1)^{n-1} = 1\). Thus, the sum of the first 40 terms of this alternating series is zero because each positive term is canceled by the next negative term. Hence: \[ \sum_{n=1}^{40} (-1)^{n-1} = 0 \] Therefore, the total sum is: \[ S_{40} = 820 + 0 = 820 \]

Thus, the sum of the first 40 terms of the series is 820.