If` y=3 x + 6 x^(2) + 10 x^(3) +…` then x =

Options

A. `(4)/(3) - (1*4)/(3^(2)*2) y^(2) + (1*4*7)/(3^(2)*3) y^(3)...`
B. `(4)/(3) + (1*4)/(3^(2)*2) y^(2) - (1*4*7)/(3^(2)*3) y^(3)...`
C. `(4)/(3) + (1*4)/(3^(2)*2) y^(2) + (1*4*7)/(3^(2)*3) y^(3)...`
D. none of these

Answer

To solve the problem, we start with the given series:

\[ y = 3x + 6x^2 + 10x^3 + \ldots \]

### Step 1: Identify the Pattern
The coefficients of the series \(3, 6, 10, \ldots\) resemble the triangular numbers. The \(n\)-th triangular number can be expressed as:

\[ T_n = \frac{n(n+1)}{2} \]

For \(n=1\), \(T_1 = 1\); for \(n=2\), \(T_2 = 3\); for \(n=3\), \(T_3 = 6\); for \(n=4\), \(T_4 = 10\). Thus, the coefficients can be expressed as:

- Coefficient of \(x^1\) is \(T_2 = 3\)
- Coefficient of \(x^2\) is \(T_3 = 6\)
- Coefficient of \(x^3\) is \(T_4 = 10\)

### Step 2: Relate the Series to a Binomial Expansion
We know that the series can be related to the binomial expansion of:

\[ (1-x)^{-n} \]

In particular, we can use the expansion for \(n=3\):

\[ (1-x)^{-3} = \sum_{k=0}^{\infty} \binom{k+2}{2} x^k = 1 + 3x + 6x^2 + 10x^3 + \ldots \]

### Step 3: Set Up the Equation
From the expansion, we can write:

\[ y = (1-x)^{-3} - 1 \]

This gives us:

\[ 1 - x = (y + 1)^{-1/3} \]

### Step 4: Solve for \(x\)
Rearranging the equation gives:

\[ x = 1 - (y + 1)^{-1/3} \]

### Step 5: Final Expression
To express \(x\) in terms of \(y\):

\[ x = 1 - \frac{1}{(y + 1)^{1/3}} \]

### Conclusion
Thus, the value of \(x\) in terms of \(y\) is:

\[ x = 1 - (y + 1)^{-1/3} \]