Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Binomial Theorem Solutions Exercise 6(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Binomial Theorem Solutions Exercise 6(a)

I.

Question 1.
Expand the following using the binomial theorem.
(i) (4x + 5y)7
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) I Q1(i)

(ii) \(\left(\frac{2}{3} x+\frac{7}{4} y\right)^5\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) I Q1(ii)

(iii) \(\left(\frac{2 p}{5}-\frac{3 q}{7}\right)^6\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) I Q1(iii)
\(\sum_{r=0}^6(-1)^{r \cdot 6} C_r\left(\frac{2 p}{5}\right)^{6-r}\left(\frac{3 q}{7}\right)^r\)

(iv) (3 + x – x2)4
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) I Q1(iv)

Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a)

Question 2.
Write down and simplify
(i) 6th term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\)
Solution:
6th term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\)
The general term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\) is
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) I Q2(i)

(ii) 7th term in (3x – 4y)10
Solution:
7th term in (3x – 4y)10
The general term in (3x – 4y)10 is
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) I Q2(ii)

(iii) 10th term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\)
Solution:
10th term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\)
General term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\) is
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) I Q2(iii)
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) I Q2(iii).1

(iv) rth term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) (1 ≤ r ≤ 9)
Solution:
rth term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\)
The general term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) is
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) I Q2(iv)

Question 3.
Find the number of terms in the expansion of
(i) \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\)
Solution:
The number of terms in (x + a)n is (n + 1), where n is a positive integer.
Hence number of terms in \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\) are 9 + 1 = 10

(ii) (3p + 4q)14
Solution:
Number of terms in (3p + 4q)14 are 14 + 1 = 15

(iii) (2x + 3y + z)7
Solution:
Number of terms in (a + b + c)n are \(\frac{(n+1)(n+2)}{2}\), where n is a positive integer.
Hence number of terms in (2x + 3y + z)7 are = \(\frac{(7+1)(7+2)}{2}=\frac{8 \times 9}{2}\) = 36

Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a)

Question 4.
Find the number of terms with non-zero coefficients in (4x – 7y)49 + (4x + 7y)49.
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) I Q4
∴ The number of terms with non-zero coefficient in (4x – 7y)49 + (4x + 7y)49 is 25.

Question 5.
Find the sum of the last 20 coefficients in the expansions of (1 + x)39.
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) I Q5
∴ The sum of the last 20 coefficients in the expansion of (1 + x)39 is 238.

Question 6.
If A and B are coefficients of xn in the expansion of (1 + x)2n and (1 + x)2n-1 respectively, then find the value of \(\frac{A}{B}\)
Solution:
Given A and B are the coefficient of xn in the expansion of (1 + x)2n and (1 + x)2n-1 respectively.
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) I Q6

II.

Question 1.
Find the coefficient of
(i) x-6 in \(\left(3 x-\frac{4}{x}\right)^{10}\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q1(i)
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q1(i).1

(ii) x11 in \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q1(ii)

(iii) x2 in \(\left(7 x^3-\frac{2}{x^2}\right)^9\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q1(iii)
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q1(iii).1

(iv) x-7 in \(\left(\frac{2 x^2}{3}-\frac{5}{4 x^5}\right)^7\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q1(iv)

Question 2.
Find the term independent of x in the expansion of
(i) \(\left(\frac{\sqrt{x}}{3}-\frac{4}{x^2}\right)^{10}\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q2(i)

(ii) \(\left(\frac{3}{\sqrt[3]{x}}+5 \sqrt{x}\right)^{25}\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q2(ii)

(iii) \(\left(4 x^3+\frac{7}{x^2}\right)^{14}\)
Solution:
The general term in \(\left(4 x^3+\frac{7}{x^2}\right)^{14}\) is
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q2(iii)
For term independent of x,
put 42 – 5r = 0
⇒ r = \(\frac{42}{5}\) which is not an integer.
Hence term independent of x in the given expansion is zero.

(iv) \(\left(\frac{2 x^2}{5}+\frac{15}{4 x}\right)^9\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q2(iv)

Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a)

Question 3.
Find the middle term(s) in the expansion of
(i) \(\left(\frac{3 x}{7}-2 y\right)^{10}\)
Solution:
The middle term in (x + a)n when n is even and is \(\frac{T_{n+1}}{2}\), when n is odd, we have two middle terms, i.e., \(\frac{T_{n+1}}{2}\) and \(\frac{T_{n+3}}{2}\)
∵ n = 10 is even,
we have only one middle term (i.e.,) \(\frac{10}{2}\) + 1 = 6th term.
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q3(i)

(ii) \(\left(4 a+\frac{3}{2} b\right)^{11}\)
Solution:
Here n = 11 is an odd integer,
we have two middle terms, i.e., \(\frac{n+1}{2}\) and \(\frac{n+3}{2}\) terms
= 6th and 7th terms are middle terms.
T6 in \(\left(4 a+\frac{3}{2} b\right)^{11}\) is \({ }^{11} C_5(4 a)^6\left(\frac{3}{2} b\right)^5\)
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q3(ii)

(iii) (4x2 + 5x3)17
Solution:
(4x2 + 5x3)17 = [x2(4 + 5x)]17 = x34(4 + 5x)17 ……..(1)
Consider (4 + 5x)17
∵ n = 17 is an odd positive integer, we have two middle terms.
They are \(\left(\frac{17+1}{2}\right)^{\text {th }}\) and \(\left(\frac{17+3}{2}\right)^{\text {th }}\) (i.e.,) 9th and 10th terms are middle terms.
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q3(iii)

(iv) \(\left(\frac{3}{a^3}+5 a^4\right)^{20}\)
Solution:
Here n = 20 is an even positive integer, we have only one middle term
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q3(iv)

Question 4.
Find the numerically greatest term(s) in the expansion of
(i) (4 + 3x)15 when x = \(\frac{7}{2}\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q4(i)
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q4(i).1

(ii) (3x + 5y)12 when x = \(\frac{1}{2}\), y = \(\frac{4}{3}\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q4(ii)
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q4(ii).1
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q4(ii).2

(iii) (4a – 6b)13 when a = 3, b = 5
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q4(iii)
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q4(iii).1
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q4(iii).2
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q4(iii).3

(iv) (3 + 7x)n when x = \(\frac{4}{5}\), n = 15
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q4(iv)
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q4(iv).1

Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a)

Question 5.
Prove the following.
(i) 2 . C0 + 5 . C1 + 8 . C2 + ……… + (3n+2) . Cn = (3n + 4) . 2n-1
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q5(i)

(ii) C0 – 4 . C1 + 7 . C2 – 10 . C3 + ……… = 0, if n is an even positive integer.
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q5(ii)

(iii) \(\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\frac{C_7}{8}+\ldots \ldots=\frac{2^n-1}{n+1}\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q5(iii)

(iv) \(C_0+\frac{3}{2} \cdot C_1+\frac{9}{3} \cdot C_2+\frac{27}{4} \cdot C_3\) + ……… + \(\frac{3^n}{n+1} \cdot C_n=\frac{4^{n+1}-1}{3(n+1)}\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q5(iv)
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q5(iv).1

(v) C0 + 2 . C1 + 4 . C2 + 8 . C3 + ….. + 2n . Cn = 3n
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q5(v)

Question 6.
Find the sum of the following.
(i) \(\frac{{ }^{15} C_1}{{ }^{15} C_0}+2 \frac{{ }^{15} C_2}{{ }^{15} C_1}+3 \frac{{ }^{15} C_3}{{ }^{15} C_2}\) + …….. + \(15 \frac{{ }^{15} C_{15}}{{ }^{15} C_{14}}\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q6(i)

(ii) C0 . C3 + C1 . C4 + C2 . C5 + …….. + Cn-3 . Cn
Solution:
We know that
(1 + x)n = C0 + C1 x + C2 x2 + ……. + Cn . xn ……….(1)
On replacing x by \(\frac{1}{x}\), we get
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q6(ii)

(iii) 22 . C0 + 32 . C1 + 42 . C2 + ……… + (n + 2)2 Cn
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q6(iii)

(iv) 3C0 + 6C1 + 12C2 + ……… + 3 . 2n . Cn
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q6(iv)

Question 7.
Using the binomial theorem, prove that 50n – 49n – 1 is divisible by 492 for all positive integers n.
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q7
= 492 [a positive integer]
Hence 50n – 49n – 1 is divisible by 492 for all positive integers of n.

Question 8.
Using the binomial theorem, prove that 54n + 52n – 1 is divisible by 676 for all positive integers n.
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q8
∴ 54n + 52n – 1 is divisible by 676, for all positive integers n.

Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a)

Question 9.
If (1 + x + x2)n = a0 + a1 x + a2 x2 + ……… + a2n x2n, then prove that
(i) a0 + a1 + a2 + ……… + a2n = 3n
(ii) a0 + a2 + a4 + …… + a2n = \(\frac{3^n+1}{2}\)
(iii) a1 + a3 + a5 + ……… + a2n-1 = \(\frac{3^n-1}{2}\)
(iv) a0 + a3 + a6 + a9 + ……….. = 3n-1
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q9
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q9.1
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q9.2

Question 10.
If (1 + x + x2 + x3)7 = b0 + b1x + b2x2 + ………. b21 x21, then find the value of
(i) b0 + b2 + b4 + …….. + b20
(ii) b1 + b3 + b5 + ………. + b21
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q10

Question 11.
If the coefficient of x11 and x12 in the binomial expansion of \(\left(2+\frac{8 x}{3}\right)^n\) are equal, find n.
Solution:
The general term of \(\left(2+\frac{8 x}{3}\right)^n\) is \(T_{r+1}={ }^n C_r(2)^{n-r}\left(\frac{8 x}{3}\right)^r\)
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q11

Question 12.
Find the remainder when 22013 is divided by 17.
Solution:
We know 24 = 16
The remainder when 24 is divided by 17 is 1
22013 = (24)503 . 21
∴ The remainder when 22013 is divided by 17 is (-1)503 . 2 = (-1) . 2 = -2

Question 13.
If the coefficients of (2r + 4)th term and (3r + 4)th term in the expansion of (1 + x)21 are equal, find r.
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) II Q13

III.

Question 1.
If the coefficients of x9, x10, x11 in the expansion of (1 + x)n are in A.P., then prove that n2 – 41n + 398 = 0.
Solution:
The coefficients of x9, x10, x11 in (1 + x)n are
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q1
⇒ (n – 9) (n – 21) = 11(n – 19)
⇒ n2 – 9n – 21n + 189 = 11n – 209
⇒ n2 – 41n + 398 = 0

Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a)

Question 2.
If 36, 84, 126 are three successive binomial coefficients in the expansion of (1 + x)n, find n.
Solution:
Let nCr-1, nCr, nCr+1 are three successive binomial coefficients in (1 + x)n.
Then nCr-1 = 36; nCr = 84 and nCr+1 = 126
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q2

Question 3.
If the 2nd, 3rd and 4th terms in the expansion of (a + x)n are respectively 240, 720, 1080, find a, x, n.
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q3
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q3.1

Question 4.
If the coefficients of rth, (r + 1)th and (r + 2)nd terms in the expansion of (1 + x)n are in A.P. then show that n2 – (4r + 1)n + 4r2 – 2 = 0.
Solution:
Coefficient of Tr = nCr-1
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q4

Question 5.
Find the sum of the coefficients of x32 and x-18 in the expansion of \(\left(2 x^3-\frac{3}{x^2}\right)^{14}\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q5
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q5.1

Question 6.
If P and Q are the sums of odd terms and the sum of even terms respectively in the expansion of (x + a)n then prove that
(i) P2 – Q2 = (x2 – a2)n
(ii) 4PQ = (x + a)2n – (x – a)2n
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q6

Question 7.
If the coefficients of 4 consecutive terms in the expansion of (1 + x)n are a1, a2, a3, a4 respectively, then show that \(\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=\frac{2 a_2}{a_2+a_3}\)
Solution:
Given a1, a2, a3, a4 are the coefficients of 4 consecutive terms in (1 + x)n respectively.
Let a1 = nCr-1, a2 = nCr, a3 = nCr+1, a4 = nCr+2
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q7
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q7.1

Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a)

Question 8.
Prove that (2nC0)2 – (2nC1)2 + (2nC2)2 – (2nC3)2 + ……… + (2nC2n)2 = (-1)n 2nCn
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q8
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q8.1

Question 9.
Prove that (C0 + C1)(C1 + C2)(C2 + C3) ………… (Cn-1 + Cn) = \(\frac{(n+1)^n}{n !}\) . C0 . C1 . C2 ……… Cn
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q9

Question 10.
Find the term independent of x in \((1+3 x)^n\left(1+\frac{1}{3 x}\right)^n\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q10

Question 11.
Show that the middle term in the expansion of (1 + x)2n is \(\frac{1.3 .5 \ldots(2 n-1)}{n !}(2 x)^n\)
Solution:
The expansion of (1 + x)2n contains (2n + 1) terms.
middle term = 2nCn xn
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q11

Question 12.
If (1 + 3x – 2x2)10 = a0 + a1x + a2x2 + …….. + a20 x20 then prove that
(i) a0 + a1 + a2 + ……… + a20 = 210
(ii) a0 – a1 + a2 – a3 + ……….. + a20 = 410
Solution:
(1 + 3x – 2x2)10 = a0 + a1x + a2x2 + ……… + a20 x20
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q12

Question 13.
If (3√3 + 5)2n+1 = x and f = x – [x] where ([x] is the integral part of x), find the value of x.f.
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q13

Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a)

Question 14.
If R, n are positive integers, n is odd, 0 < F < 1 and if (5√5 + 11)n = R + F, then prove that
(i) R is an even integer and
(ii) (R + F) . F = 4n
Solution:
(i) Since R, n are positive integers, 0 < F < 1 and (5√5 + 11)n = R + F
Let (5√5 – 11)n = f
Now, 11 < 5√5 < 12
⇒ 0 < 5√5 – 11 < 1
⇒ 0 < (5√5 – 11)n < 1
⇒ 0 < f < 1
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q14
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q14.1

Question 15.
If I, n are positive integers, 0 < f < 1 and if (7 + 4√3 )n = I + f, then show that
(i) I is an odd integer and
(ii) (I + f) (1 – f) = 1
Solution:
Given I, n are positive integers and
(7 + 4√3)n = I + f, 0 < f < 1
Let 7 – 4√3 = F
Now 6 < 4√3 < 7
⇒ -6 > -4√3 > -7
⇒ 1 > 7 – 4√3 > 0
⇒ 0 < (7 – 4√3)n < 1
∴ 0 < F < 1
I + f + F = (7 + 4√3)n + (7 – 4√3)n
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q15
= 2k where k is an integer.
∴ I + f + F is an even integer.
⇒ f + F is an integer since I is an integer.
But 0 < f < 1 and 0 < F < 1
⇒ 0 < f + F < 2
∴ f + F = 1 ………..(1)
⇒ I + 1 is an even integer.
∴ I is an odd integer.
(I + f) (I – f) = (I + f) F …..[By (1)]
= (7 + 4√3)n (7 – 4√3)n
= [(7 + 4√3) (7 – 4√3)]n
= (49 – 48)n
= 1

Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a)

Question 16.
If n is a positive integer, prove that \(\sum_{r=1}^n r^3\left(\frac{{ }^n C_r}{{ }^n C_{r-1}}\right)^2=\frac{(n)(n+1)^2(n+2)}{12}\)
Solution:
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q16
Inter 2nd Year Maths 2A Binomial Theorem Solutions Ex 6(a) III Q16.1

Question 17.
Find the number of irrational terms in the expansion of (51/6 + 21/8)100.
Solution:
General term
Tr+1 = \({ }^{100} C_r\left(5^{1 / 6}\right)^{100-r}\left(2^{1 / 8}\right)^r\) = \({ }^{100} C_r 5^{\frac{100-r}{6}} \cdot 2^{\frac{r}{8}}\)
\(\frac{100-r}{6}\) is an integer in the span
or 0 ≤ r ≤ 100 if r = 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, 70, 76, 82, 88, 94, 100
\(\frac{r}{8}\) is an integer in the span of 0 ≤ r ≤ 100
if r = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, \(\frac{100-r}{6}\), \(\frac{r}{8}\) both an integers
If r = 16, 40, 64, 88
∴ The number of rational terms in the expansion of (51/6 + 21/8)r is 4.
∴ The number of irrational terms in the expansion of (51/6 + 21/8)r is 101 – 4 = 97 terms.