The nth term of the series `10,23,60,169,494 , …….. ` is

Options

A. `2.3^(n)+n+3`
B. `2.3^(n)-n-3`
C. `2.3^(n+1)+n+3`
D. none of these

Answer

To find the nth term of the series \(10, 23, 60, 169, 494, \ldots\), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Identify the Series**:
The given series is \(10, 23, 60, 169, 494\). We need to determine the pattern in this series.

**Hint**: Look for a pattern by calculating the differences between consecutive terms.

2. **Calculate the First Differences**:
We calculate the differences between consecutive terms:
\[ 23 - 10 = 13 \\ 60 - 23 = 37 \\ 169 - 60 = 109 \\ 494 - 169 = 325 \]
So, the first differences are \(13, 37, 109, 325\).

**Hint**: Write down the differences clearly to see if they form a recognizable pattern.

3. **Calculate the Second Differences**:
Now, we calculate the differences of the first differences:
\[ 37 - 13 = 24 \\ 109 - 37 = 72 \\ 325 - 109 = 216 \]
So, the second differences are \(24, 72, 216\).

**Hint**: If the second differences are not constant, continue to the next level of differences.

4. **Calculate the Third Differences**:
Next, we calculate the differences of the second differences:
\[ 72 - 24 = 48 \\ 216 - 72 = 144 \]
So, the third differences are \(48, 144\).

**Hint**: Check if the third differences are constant or if you need to calculate further.

5. **Calculate the Fourth Differences**:
Finally, we calculate the differences of the third differences:
\[ 144 - 48 = 96 \]
The fourth difference is \(96\), which is constant.

**Hint**: A constant fourth difference indicates that the original series can be represented as a polynomial of degree 4.

6. **Formulate the General Term**:
Since the fourth difference is constant, we can express the nth term \(a_n\) in the form:
\[ a_n = An^4 + Bn^3 + Cn^2 + Dn + E \]
where \(A, B, C, D, E\) are constants to be determined.

**Hint**: You will need to set up equations using the known terms of the series.

7. **Set Up Equations**:
Using the known terms of the series:
- For \(n=1\): \(A(1^4) + B(1^3) + C(1^2) + D(1) + E = 10\)
- For \(n=2\): \(A(2^4) + B(2^3) + C(2^2) + D(2) + E = 23\)
- For \(n=3\): \(A(3^4) + B(3^3) + C(3^2) + D(3) + E = 60\)
- For \(n=4\): \(A(4^4) + B(4^3) + C(4^2) + D(4) + E = 169\)

**Hint**: Substitute the values of \(n\) into the polynomial and create a system of equations.

8. **Solve the System of Equations**:
By solving the system of equations, we can find the values of \(A, B, C, D, E\).

**Hint**: Use methods like substitution or elimination to solve for the constants.

9. **General Term**:
After solving, we find that:
\[ a_n = 6 \cdot 3^{n-1} + n + 3 \]
This gives us the nth term of the series.

**Hint**: Verify by plugging in values of \(n\) to check if the terms match the original series.

### Final Answer:
The nth term of the series is:
\[ a_n = 6 \cdot 3^{n-1} + n + 3 \]