Find the sixth term of the expansion of `(y^(1//2) + x^(1//3) )^(n)`, if the binomial coefficient of the third term from the end is `45`.

Answer

To find the sixth term of the expansion of \((y^{1/2} + x^{1/3})^n\) given that the binomial coefficient of the third term from the end is \(45\), we can follow these steps:

### Step 1: Understand the Binomial Expansion
The general term \(T_{r+1}\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = y^{1/2}\) and \(b = x^{1/3}\).

### Step 2: Identify the Third Term from the End
The third term from the end corresponds to \(T_{n-2}\) (since the last term is \(T_n\), the second last is \(T_{n-1}\), and the third last is \(T_{n-2}\)).

### Step 3: Set Up the Coefficient Equation
The coefficient of the third term from the end is given as \(45\):
\[ \binom{n}{n-2} = \binom{n}{2} = 45 \]
This implies:
\[ \frac{n(n-1)}{2} = 45 \]

### Step 4: Solve for \(n\)
Multiplying both sides by \(2\):
\[ n(n-1) = 90 \]
Rearranging gives:
\[ n^2 - n - 90 = 0 \]
Now we can factor this quadratic equation:
\[ (n - 10)(n + 9) = 0 \]
Thus, \(n = 10\) or \(n = -9\). Since \(n\) must be a natural number, we have:
\[ n = 10 \]

### Step 5: Find the Sixth Term
Now that we have \(n = 10\), we can find the sixth term \(T_6\):
\[ T_6 = \binom{10}{5} (y^{1/2})^{10-5} (x^{1/3})^5 \]
This simplifies to:
\[ T_6 = \binom{10}{5} (y^{1/2})^5 (x^{1/3})^5 \]
Calculating \(\binom{10}{5}\):
\[ \binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]
Thus:
\[ T_6 = 252 (y^{5/2}) (x^{5/3}) \]

### Final Answer
The sixth term of the expansion is:
\[ 252 y^{5/2} x^{5/3} \]