The expression
`N = 1/(1 + 1 xx 1) + 1/(2 + 2 xx 2) + 1/(3 + 3 xx 3) + 1/(4 + 4 xx 4) + …. 1/(n + n xx n')` where n = 100, then value of N is:

Options

A. `100/101`
B. `99/100`
C. 9900
D. none of these

Answer

To solve the expression

\[ N = \frac{1}{1 + 1 \times 1} + \frac{1}{2 + 2 \times 2} + \frac{1}{3 + 3 \times 3} + \frac{1}{4 + 4 \times 4} + \ldots + \frac{1}{n + n \times n} \]

where \( n = 100 \), we can simplify each term in the series.

### Step 1: Simplify each term

Each term in the series can be expressed as:

\[ \frac{1}{k + k \times k} = \frac{1}{k + k^2} = \frac{1}{k(1 + k)} \]

Thus, we can rewrite \( N \) as:

\[ N = \sum_{k=1}^{100} \frac{1}{k(1 + k)} \]

### Step 2: Decompose the fraction

We can use partial fraction decomposition on the term \( \frac{1}{k(1 + k)} \):

\[ \frac{1}{k(1 + k)} = \frac{1}{k} - \frac{1}{1 + k} \]

### Step 3: Rewrite the sum

Now substituting back into the sum, we have:

\[ N = \sum_{k=1}^{100} \left( \frac{1}{k} - \frac{1}{1 + k} \right) \]

### Step 4: Expand the sum

This can be expanded as:

\[ N = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \ldots + \left( \frac{1}{100} - \frac{1}{101} \right) \]

### Step 5: Notice the cancellation

In this series, most terms will cancel out:

\[ N = 1 - \frac{1}{101} \]

### Step 6: Calculate the final value

Thus, we can simplify this to:

\[ N = 1 - \frac{1}{101} = \frac{101}{101} - \frac{1}{101} = \frac{100}{101} \]

### Final Answer

So, the value of \( N \) is:

\[ N = \frac{100}{101} \]

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