The number of integral terms in the expansion of (5^(1/2) + 7^(1/8))^1024 is

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q17 – AP-EAMCET 2026

← Back

Home Maths Binomial TheoremQuestion #35

The number of integral terms in the expansion of

(5^{\frac{1}{2}} + 7^{\frac{1}{8}})^{1024}

Clue 1 of 3: General Term in Binomial Expansion

Recall the formula for the general term $T_{r + 1}$ in the binomial expansion of $(a + b)^{n}$ .

🧠 Full Solution Path

Step 1: Write the General Term of the Expansion✦ Active

The given expression is $(5^{\frac{1}{2}} + 7^{\frac{1}{8}})^{1024}$ . We need to find the number of integral terms in its expansion. The general term in the expansion of $(a + b)^{n}$ is given by $T_{r + 1} = (\binom{n}{r}) a^{n - r} b^{r}$ . Here, $a = 5^{\frac{1}{2}}$ , $b = 7^{\frac{1}{8}}$ , and $n = 1024$ . Substituting these values into the general term formula:

T_{r + 1} = (\binom{1024}{r}) (5^{\frac{1}{2}})^{1024 - r} (7^{\frac{1}{8}})^{r}

Simplify the exponents:

T_{r + 1} = (\binom{1024}{r}) 5^{\frac{1024 - r}{2}} 7^{\frac{r}{8}}

💡 Teacher's Secret Hint

Remember that $r$ is an integer ranging from $0$ to $n$ inclusive.

Step 2: Set Conditions for Integral Terms○ Expand

For the term $T_{r + 1}$ to be an integer, the powers of the prime numbers 5 and 7 must be non-negative integers. Also, the binomial coefficient $(\binom{1024}{r})$ is always an integer. Thus, we require:

1. $\frac{1024 - r}{2}$ must be an integer.

2. $\frac{r}{8}$ must be an integer.

💡 Teacher's Secret Hint

If an exponent is not an integer, the term will involve roots, making it non-integral (unless the base is a perfect power of the root, which is not the case for 5 and 7 here).

Step 3: Determine the Possible Values of r○ Expand

From condition (2), for $\frac{r}{8}$ to be an integer, $r$ must be a multiple of 8. So, $r = 8 k$ for some integer $k$ .

From condition (1), for $\frac{1024 - r}{2}$ to be an integer, $1024 - r$ must be an even number. Since 1024 is an even number, $r$ must also be an even number. This condition is automatically satisfied if $r$ is a multiple of 8 (as $8 k$ is always even).

The index $r$ in the binomial expansion ranges from $0$ to $n$ . Here, $n = 1024$ . So, $0 \leq r \leq 1024$ .

Substituting $r = 8 k$ into the inequality:

0 \leq 8 k \leq 1024

Divide by 8:

0 \leq k \leq \frac{1024}{8}

0 \leq k \leq 128

Step 4: Count the Number of Integral Terms○ Expand

The possible integer values for $k$ are $0, 1, 2, \dots, 128$ . To find the number of these values, we use the formula (last value - first value + 1).

Number of terms = $128 - 0 + 1 = 129$ .

Thus, there are 129 integral terms in the expansion.

💡 Teacher's Secret Hint

Be careful not to forget the $r = 0$ term when counting, as it is often a source of error (i.e., $128$ vs $129$ ). The number of integers from $A$ to $B$ (inclusive) is $B - A + 1$ .

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 17 – AP-EAMCET 2026

Clue 1 of 3: General Term in Binomial Expansion

🧠 Full Solution Path