`(2^(1/2)+5^(1/8))^n` has 33 integral terms find least value of n is-

Options

A. 256
B. 257
C. 258
D. 259

Answer

To solve the problem `(2^(1/2) + 5^(1/8))^n` having 33 integral terms, we will follow these steps:

### Step 1: Understand the Binomial Expansion
The expression can be expanded using the binomial theorem:
\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]
where \( a = 2^{1/2} \) and \( b = 5^{1/8} \).

### Step 2: Identify the General Term
The general term \( T_r \) in the expansion is given by:
\[ T_r = \binom{n}{r} (2^{1/2})^{n-r} (5^{1/8})^r = \binom{n}{r} 2^{(n-r)/2} 5^{r/8} \]

### Step 3: Conditions for Integral Terms
For \( T_r \) to be an integer, both \( 2^{(n-r)/2} \) and \( 5^{r/8} \) must be integers. This leads to two conditions:
1. \( \frac{n-r}{2} \) must be an integer, which implies \( n - r \) is even.
2. \( \frac{r}{8} \) must be an integer, which implies \( r \) is a multiple of 8.

### Step 4: Express Conditions in Terms of r
Let \( r = 8k \) for \( k = 0, 1, 2, \ldots \). Then, we can express \( n - r \) as:
\[ n - 8k \text{ must be even} \]
This means \( n \) must also be even since \( 8k \) is always even.

### Step 5: Determine the Range of k
The values of \( r \) must satisfy:
\[ 0 \leq r \leq n \implies 0 \leq 8k \leq n \implies 0 \leq k \leq \frac{n}{8} \]
The number of valid \( k \) values is:
\[ k = 0, 1, 2, \ldots, \left\lfloor \frac{n}{8} \right\rfloor \]
Thus, the total number of integral terms is:
\[ \left\lfloor \frac{n}{8} \right\rfloor + 1 \]

### Step 6: Set Up the Equation
We know the total number of integral terms is 33:
\[ \left\lfloor \frac{n}{8} \right\rfloor + 1 = 33 \]
This simplifies to:
\[ \left\lfloor \frac{n}{8} \right\rfloor = 32 \]

### Step 7: Solve for n
This implies:
\[ 32 \leq \frac{n}{8} < 33 \implies 256 \leq n < 264 \]
The least integer value for \( n \) that satisfies this inequality is:
\[ n = 256 \]

### Conclusion
Thus, the least value of \( n \) for which the expression has 33 integral terms is:
\[ \boxed{256} \]