Question

The number of integral terms in the expansion of $ \left( 5^{\frac{1}{2}} + 7^{\frac{1}{8}} \right)^{1016} $ is:

• A. 130
• B. 128
• C. 127
• D. 129

Hint:

For binomial expansions involving fractional exponents, ensure that the exponents of the terms are integers for the terms to be integral. This can be done by ensuring that the powers of the terms satisfy the divisibility conditions.

Answer 1

The Correct Option is B

To determine the number of integral terms in the expansion of the expression \( \left( 5^{\frac{1}{2}} + 7^{\frac{1}{8}} \right)^{1016} \), we will use the binomial theorem. The general term in the expansion of \( (a + b)^n \) is given by:

\(T_{k+1} = \binom{n}{k} a^{n-k} b^k\)

For the specific case of this expression:

• Let \( a = 5^{\frac{1}{2}} \) and \( b = 7^{\frac{1}{8}} \).
• The general term becomes:
  \(T_{k+1} = \binom{1016}{k} (5^{\frac{1}{2}})^{1016-k} (7^{\frac{1}{8}})^k\)
• This can be simplified to:
  \(T_{k+1} = \binom{1016}{k} 5^{\frac{1016-k}{2}} 7^{\frac{k}{8}}\)

For the term to be integral, the exponents of both 5 and 7 should be integers. This implies:

• \(\frac{1016-k}{2} = \text{integer} \Rightarrow 1016-k \text{ must be even}\)
• \(\frac{k}{8} = \text{integer} \Rightarrow k \text{ must be a multiple of 8}\)

Such a value of \( k \) must satisfy both conditions: \( k = 0, 8, 16, \dots \) such that \( 1016-k \) is even. Given that \( 1016 \) is already even, all even \( k \) derived from multiples of 8 will satisfy the first condition.

The sequence for \( k \) as a multiple of 8 is:

\( k = 0, 8, 16, \dots, 1016 \)

These values form an arithmetic sequence with the first term as 0, common difference of 8, and last term 1016. To find the number of terms, we use the formula for the nth-term of an arithmetic sequence:

\(k = a + (n-1)d \Rightarrow 1016 = 0 + (n-1) \cdot 8\)

Solving for \( n \):

\(1016 = 8(n-1) \Rightarrow n-1 = \frac{1016}{8} = 127 \Rightarrow n = 128\)

Hence, the number of integral terms is 128.

Answer 2

Step 1: General Form of the Expansion The given expression is of the form \( (a + b)^n \), where: \[ a = 5^{\frac{1}{2}}, \quad b = 7^{\frac{1}{8}}, \quad n = 1016. \] The general term in the binomial expansion of \( (a + b)^n \) is: \[ T_r = \binom{n}{r} a^{n-r} b^r. \] Substituting \( a = 5^{\frac{1}{2}} \) and \( b = 7^{\frac{1}{8}} \), we get the general term: \[ T_r = \binom{1016}{r} \left( 5^{\frac{1}{2}} \right)^{1016-r} \left( 7^{\frac{1}{8}} \right)^r = \binom{1016}{r} \cdot 5^{\frac{1016 - r}{2}} \cdot 7^{\frac{r}{8}}. \] Thus, the general term is: \[ T_r = \binom{1016}{r} \cdot 5^{\frac{1016 - r}{2}} \cdot 7^{\frac{r}{8}}. \]

Step 2: Identifying the Conditions for Integral Terms For the term \( T_r \) to be an integer, both \( 5^{\frac{1016 - r}{2}} \) and \( 7^{\frac{r}{8}} \) should be integers.
This means that the exponents of 5 and 7 must be integers. For \( 5^{\frac{1016 - r}{2}} \) to be an integer, \( \frac{1016 - r}{2} \) must be an integer, implying that \( 1016 - r \) must be even.
Therefore, \( r \) must be even. For \( 7^{\frac{r}{8}} \) to be an integer, \( \frac{r}{8} \) must be an integer, implying that \( r \) must be a multiple of 8.

Step 3: Finding the Range of \( r \) Since \( r \) must be an even number and a multiple of 8, \( r \) must be a multiple of 8.
The possible values of \( r \) are given by the set of multiples of 8, i.e., \( r = 0, 8, 16, \dots, 1016 \). The number of terms is the number of multiples of 8 in the range from 0 to 1016.
The multiples of 8 in this range are \( 0, 8, 16, \dots, 1016 \), which form an arithmetic progression with the first term 0, the common difference 8, and the last term 1016. The number of terms in this progression is: \[ \frac{1016 - 0}{8} + 1 = 128. \] Thus, there are 128 integral terms in the expansion.