If `x = 1 + 3a + 6a^2 + 10 a^3 + ……" to " oo` terms `|a| lt 1, y = 1+4a+10a^2 + 20a^3 + ….." to " oo` terms, `|a| lt 1`, then x:y

Options

A. `(1 - a):1`
B. `1:(1 - a)`
C. `(1+a):1`
D. `1:(1+a)`

Answer

To solve the problem, we need to find the ratio \( \frac{x}{y} \) where:

\[ x = 1 + 3a + 6a^2 + 10a^3 + \ldots \]
\[ y = 1 + 4a + 10a^2 + 20a^3 + \ldots \]

### Step 1: Finding \( x \)

We start with the series for \( x \):
\[ x = 1 + 3a + 6a^2 + 10a^3 + \ldots \]

Notice that the coefficients \( 1, 3, 6, 10, \ldots \) are the triangular numbers, which can be expressed as \( T_n = \frac{n(n+1)}{2} \). Thus, we can rewrite \( x \) as:
\[ x = \sum_{n=0}^{\infty} T_n a^n \]

### Step 2: Relating \( x \) to a simpler series

We can express \( T_n \) in terms of sums:
\[ T_n = \sum_{k=1}^{n} k \]

So, we can express \( x \) as:
\[ x = \sum_{n=0}^{\infty} \left( \sum_{k=1}^{n} k \right) a^n \]

Now, we can manipulate this series. Multiply \( x \) by \( a \):
\[ ax = a + 3a^2 + 6a^3 + 10a^4 + \ldots \]

### Step 3: Subtracting the two series

Subtract \( ax \) from \( x \):
\[ x - ax = 1 + (3a - a) + (6a^2 - 3a^2) + (10a^3 - 6a^3) + \ldots \]
\[ x(1 - a) = 1 + 2a + 3a^2 + 4a^3 + \ldots \]

The right-hand side is a known series:
\[ 1 + 2a + 3a^2 + 4a^3 + \ldots = \frac{1}{(1-a)^2} \]

Thus, we have:
\[ x(1 - a) = \frac{1}{(1-a)^2} \]
\[ x = \frac{1}{(1-a)^3} \]

### Step 4: Finding \( y \)

Now we analyze \( y \):
\[ y = 1 + 4a + 10a^2 + 20a^3 + \ldots \]

The coefficients \( 1, 4, 10, 20, \ldots \) can be expressed as \( \frac{n(n+1)}{2} + n \), which leads us to:
\[ y = \sum_{n=0}^{\infty} \left( \frac{n(n+1)}{2} + n \right) a^n \]

### Step 5: Relating \( y \) to a simpler series

Similar to \( x \), we can express \( y \) as:
\[ y = \sum_{n=0}^{\infty} T_n a^n + \sum_{n=0}^{\infty} n a^n \]

The second series \( \sum_{n=0}^{\infty} n a^n = \frac{a}{(1-a)^2} \).

Thus, we have:
\[ y = \frac{1}{(1-a)^3} + \frac{a}{(1-a)^2} \]

### Step 6: Finding the ratio \( \frac{x}{y} \)

Now we can find the ratio:
\[ \frac{x}{y} = \frac{\frac{1}{(1-a)^3}}{\frac{1}{(1-a)^3} + \frac{a}{(1-a)^2}} \]

Simplifying this gives:
\[ \frac{x}{y} = \frac{1}{1 + a(1-a)} \]
\[ = \frac{1}{1 - a + a^2} \]

### Final Result

Thus, the ratio \( x:y \) is:
\[ \frac{x}{y} = \frac{1}{1-a} \]