Evaluate `:1 + 3x + 6x ^(2) + 10 x ^(3)+…….` upto infinite term, whre `|x| lt 1.`

Answer

To evaluate the series \( S = 1 + 3x + 6x^2 + 10x^3 + \ldots \) up to infinite terms, where \( |x| < 1 \), we can follow these steps:

### Step 1: Identify the series
The series can be expressed as:
\[ S = 1 + 3x + 6x^2 + 10x^3 + \ldots \]

### Step 2: Write the series in terms of \( S \)
Let’s denote the series as \( S \):
\[ S = 1 + 3x + 6x^2 + 10x^3 + \ldots \]

### Step 3: Multiply the series by \( x \)
Now, multiply the entire series \( S \) by \( x \):
\[ xS = x + 3x^2 + 6x^3 + 10x^4 + \ldots \]

### Step 4: Subtract the two equations
Now, subtract the equation for \( xS \) from \( S \):
\[ S - xS = (1 + 3x + 6x^2 + 10x^3 + \ldots) - (x + 3x^2 + 6x^3 + 10x^4 + \ldots) \]
This simplifies to:
\[ S(1 - x) = 1 + (3x - x) + (6x^2 - 3x^2) + (10x^3 - 6x^3) + \ldots \]
\[ S(1 - x) = 1 + 2x + 3x^2 + 4x^3 + \ldots \]

### Step 5: Recognize the new series
The series \( 1 + 2x + 3x^2 + 4x^3 + \ldots \) can be recognized as another series. We can denote it as \( T \):
\[ T = 1 + 2x + 3x^2 + 4x^3 + \ldots \]

### Step 6: Multiply \( T \) by \( x \)
Now, multiply \( T \) by \( x \):
\[ xT = x + 2x^2 + 3x^3 + 4x^4 + \ldots \]

### Step 7: Subtract the two equations again
Subtract \( xT \) from \( T \):
\[ T - xT = (1 + 2x + 3x^2 + 4x^3 + \ldots) - (x + 2x^2 + 3x^3 + 4x^4 + \ldots) \]
This simplifies to:
\[ T(1 - x) = 1 + (2x - x) + (3x^2 - 2x^2) + (4x^3 - 3x^3) + \ldots \]
\[ T(1 - x) = 1 + x + x^2 + x^3 + \ldots \]

### Step 8: Recognize the new series as a geometric series
The series \( 1 + x + x^2 + x^3 + \ldots \) is a geometric series with first term \( 1 \) and common ratio \( x \). The sum of this series is:
\[ \frac{1}{1 - x} \quad \text{(for } |x| < 1\text{)} \]

### Step 9: Substitute back
Now we have:
\[ T(1 - x) = \frac{1}{1 - x} \]
Thus,
\[ T = \frac{1}{(1 - x)(1 - x)} = \frac{1}{(1 - x)^2} \]

### Step 10: Substitute \( T \) back into the equation for \( S \)
Now substitute \( T \) back into the equation for \( S \):
\[ S(1 - x) = T = \frac{1}{(1 - x)^2} \]
So,
\[ S = \frac{1}{(1 - x)^2(1 - x)} = \frac{1}{(1 - x)^3} \]

### Final Answer
Thus, the sum of the series is:
\[ S = \frac{1}{(1 - x)^3} \]