The number of integral terms in the expansion of (5^(1/2) + 7^(1/8))^1016 is

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Maths Q8 – JEE-MAIN 2025

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Home Maths Binomial TheoremQuestion #19

The number of integral terms in the expansion of

{(5^{\frac{1}{2}} + 7^{\frac{1}{8}})}^{1016}

Clue 1 of 3: Binomial Expansion

Recall the general term in the binomial expansion of $(a + b)^{n}$ .

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Step 1: Write the general term✦ Active

The given expression is ${(5^{\frac{1}{2}} + 7^{\frac{1}{8}})}^{1016}$ . The general term $T_{r + 1}$ in the expansion of $(a + b)^{n}$ is $T_{r + 1} = (\binom{n}{r}) a^{n - r} b^{r}$ . Substituting $a = 5^{\frac{1}{2}}$ , $b = 7^{\frac{1}{8}}$ , and $n = 1016$ , we get:

T_{r + 1} = (\binom{1016}{r}) {(5^{\frac{1}{2}})}^{1016 - r} {(7^{\frac{1}{8}})}^{r}

Simplifying the powers:

T_{r + 1} = (\binom{1016}{r}) 5^{\frac{1016 - r}{2}} 7^{\frac{r}{8}}

Step 2: Determine conditions for integral terms○ Expand

For $T_{r + 1}$ to be an integral term, the powers of 5 and 7 must be non-negative integers. This means:

\frac{1016 - r}{2} must be an integer and \frac{r}{8} must be an integer

From $\frac{r}{8}$ being an integer, $r$ must be a multiple of 8. So, $r = 8 k$ for some integer $k$ . From $\frac{1016 - r}{2}$ being an integer, $1016 - r$ must be an even number. Since 1016 is even, $r$ must also be even. As $r$ is a multiple of 8, it is automatically even, satisfying the second condition. The range for $r$ is $0 \leq r \leq n$ , so $0 \leq r \leq 1016$ .

Step 3: Count the number of possible values for r○ Expand

We need $r$ to be a multiple of 8, such that $0 \leq r \leq 1016$ . Let $r = 8 k$ . Substituting this into the inequality:

0 \leq 8 k \leq 1016

Dividing by 8:

0 \leq k \leq \frac{1016}{8}

0 \leq k \leq 127

The possible integer values for $k$ are $0, 1, 2, \dots, 127$ . The number of such values is $127 - 0 + 1 = 128$ . Thus, there are 128 integral terms in the expansion.

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Maths Question 8 – JEE-MAIN 2025

Clue 1 of 3: Binomial Expansion

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