The number of integral terms in the expansion of (3(1)/(2) + 5(1)/(4)) ^ 680 is equal to

Correct answer: 0 Given expansion is (3(1)/(2) + 5(1)/(4)) ^ 680 . The general term in the expansion (x + a) ^ n is T _ r + 1 = Cr n · x ^ n - r · a ^ r ⇒ T _ r + 1 = Cr 680 (3(1)/(2)) ^ 680 - r (5(1)/(4)) ^ r , 0 ≤ r ≤ 680 = Cr 680 (3)(680 - r)/(2)(5)(r)/(4) For (3)(680 - r)/(2)(5)(r)/(4) to be rational r should be a multiple of 4 . ⇒ r = 0, 4, 8, 12, . . . . . 680 ⇒ a _ n = a + (n - 1)d ⇒ 680 = 0 + (n - 1)4 ⇒ n = 171 That means r can take 171 values. Hence, the required answer is 171 .

JEE Main 202311 AprilMorning Shift

Exam: JEE Main
Year: 2023
Date: 11 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2023 · 11 April · Morning Shift · Mathematics · Binomial Theorem

The number of integral terms in the expansion of (3(1)/(2) +...

The number of integral terms in the expansion of ${\left(3\dfrac{1}{2} + 5\dfrac{1}{4}\right)}^{680} is equal to

Correct answer

Step-by-step solution

Given expansion is ${\left(3\dfrac{1}{2} + 5\dfrac{1}{4}\right)}^{680}$ .
The general term in the expansion ${\left(x + a\right)}^{n} is$ {T}_{r + 1} = Cr n \cdot {x}^{n - r} \cdot {a}^{r} $$ ⇒ {T}_{r + 1} = Cr 680 {\left(3\dfrac{1}{2}\right)}^{680 - r}{\left(5\dfrac{1}{4}\right)}^{r}, 0 \le r \le 680 $$ = Cr 680 \left(3\right)\dfrac{680 - r}{2}\left(5\right)\dfrac{r}{4} $For \left(3\right)\dfrac{680 - r}{2}\left(5\right)\dfrac{r}{4}$ to be rational $r$ should be a multiple of $4$ .
$⇒ r = 0, 4, 8, 12, . . . . . 680$
$⇒ {a}_{n} = a + \left(n - 1\right)d$
$⇒ 680 = 0 + \left(n - 1\right)4$
$⇒ n = 171$
That means $r$ can take $171$ values.
Hence, the required answer is $171$ .

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