Let `p=1+1/(sqrt(2))+1/(sqrt(3))+….+1/(sqrt(120))` and `q=1/(sqrt(2))+1/(sqrt(3))+….+1/(sqrt(121))` then

Options

A. `pgt20`
B. `qlt20`
C. `p+qlt40`
D. `p+qgt40`

Explanation

To solve the problem, we need to analyze the expressions for \( p \) and \( q \):

1. **Define the expressions**:
- \( p = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{120}} \)
- \( q = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{121}} \)

2. **Identify the function**:
- Let \( f(x) = \frac{1}{\sqrt{x}} \). This function is decreasing for \( x > 0 \).

3. **Estimate \( p \)**:
- We can estimate \( p \) using integration:
\[ p = 1 + \sum_{n=2}^{120} \frac{1}{\sqrt{n}} > \int_{1}^{121} f(x) \, dx \]
- Calculate the integral:
\[ \int f(x) \, dx = \int \frac{1}{\sqrt{x}} \, dx = 2\sqrt{x} + C \]
- Evaluating the definite integral:
\[ \int_{1}^{121} \frac{1}{\sqrt{x}} \, dx = 2\sqrt{121} - 2\sqrt{1} = 2 \times 11 - 2 \times 1 = 22 - 2 = 20 \]
- Therefore, \( p > 20 \).

4. **Estimate \( q \)**:
- Similarly, for \( q \):
\[ q = \sum_{n=2}^{121} \frac{1}{\sqrt{n}} < \int_{1}^{121} f(x) \, dx \]
- From the previous calculation, we have:
\[ q < 20 \]

5. **Combine the results**:
- From the estimates of \( p \) and \( q \):
\[ p > 20 \quad \text{and} \quad q < 20 \]
- Adding these inequalities:
\[ p + q > 20 + (q < 20) \implies p + q > 40 \]

6. **Conclusion**:
- The valid option based on our calculations is:
\[ \text{Option 4: } p + q > 40 \]