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JEE Main 20203 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
3 September
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2020 · 3 September · Morning Shift · Mathematics · Binomial Theorem

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

If the number of integral terms in the expansion of (312+518)nisexactly{\left(3\dfrac{1}{2} + 5\dfrac{1}{8}\right)}^{n} is exactly33,thentheleastvalueof, then the least value ofn$ is

Options

  1. A264
  2. B128
  3. C256
  4. D248

Correct answer

C. 256

Step-by-step solution

(312+518)n{\left(3\dfrac{1}{2} + 5\dfrac{1}{8}\right)}^{n}

General term Tr+1=Crn(312)n−r(518)r{T}_{r + 1} = Cr n {\left(3\dfrac{1}{2}\right)}^{n - r}{\left(5\dfrac{1}{8}\right)}^{r}

Tr+1=Crn3n−r25r8{T}_{r + 1} = Cr n 3\dfrac{n - r}{2}5\dfrac{r}{8}.

n−r2,r8\dfrac{n - r}{2}, \dfrac{r}{8} must be integer.

n−r2∈I⇒n−r=0,2,4,6,8.......\dfrac{n - r}{2} \in I ⇒ n - r = 0, 2, 4, 6, 8 . . . . . . .

r8∈I⇒r=0,8,16,24,32....\dfrac{r}{8} \in I ⇒ r = 0, 8, 16, 24, 32 . . . .

For rational term common powers=0,8,16,24.......= 0, 8, 16, 24 . . . . . . .

Number of integral term=33= 33

Ts=a+(s−1)d⇒T33=0+(33−1)8=256{T}_{s} = a + \left(s - 1\right)d ⇒ {T}_{33} = 0 + \left(33 - 1\right)8 = 256.

Value of n=256n = 256.

We can cross check to verify answer.

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