If 3+5+9+17+33+… to n terms `=2^(n+1)+n-2`, then nth term of LHS, is

Options

A. `3^(n)-1`
B. `2n+1`
C. `2^(n)+1`
D. `3n-1`

Answer

To find the nth term of the series given by \(3 + 5 + 9 + 17 + 33 + \ldots\) up to \(n\) terms, which is equal to \(2^{(n+1)} + n - 2\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Given Series**:
We are given that the sum of the first \(n\) terms of the series is:
\[ S_n = 2^{(n+1)} + n - 2 \]

2. **Finding the nth Term**:
The nth term of the series can be expressed in terms of the sum of the first \(n\) terms and the sum of the first \(n-1\) terms:
\[ T_n = S_n - S_{n-1} \]
where \(T_n\) is the nth term.

3. **Calculating \(S_{n-1}\)**:
We can find \(S_{n-1}\) by substituting \(n-1\) into the expression for \(S_n\):
\[ S_{n-1} = 2^{(n-1)+1} + (n-1) - 2 = 2^n + n - 3 \]

4. **Substituting into the nth Term Formula**:
Now we can substitute \(S_n\) and \(S_{n-1}\) into the formula for \(T_n\):
\[ T_n = S_n - S_{n-1} = (2^{(n+1)} + n - 2) - (2^n + n - 3) \]

5. **Simplifying the Expression**:
Simplifying the above expression:
\[ T_n = 2^{(n+1)} + n - 2 - 2^n - n + 3 \]
\[ T_n = 2^{(n+1)} - 2^n + 1 \]

6. **Factoring the Expression**:
We can factor out \(2^n\) from the first two terms:
\[ T_n = 2^n(2 - 1) + 1 = 2^n + 1 \]

7. **Final Result**:
Therefore, the nth term of the series is:
\[ T_n = 2^n + 1 \]