JEE Mathematics Binomial Theorem for Positive Integral Index MCQs Set 05

Answer: (a) 97

Explanation:
1. \( T_{r+1}={}^{100}C_r\,5^{(100-r)/8}\,2^{r/6} \) is rational when \( 8\mid(100-r) \) and \( 6\mid r \).
2. The values \( r=0,\dots,100 \) satisfying both are \( r=12,36,60,84 \), so there are 4 rational terms.
3. The expansion has 101 terms, so the number of irrational terms is \( 101-4=97 \).

Teacher's Note:
1. A term is rational only if every fractional exponent becomes an integer.
2. Subtract the rational terms from the total of n + 1 terms.

Answer: (b) 7

Explanation:
1. A general term is \( \dfrac{6!}{i!\,j!\,k!}\,(\sqrt2)^{j}(3^{1/3})^{k} \) with \( i+j+k=6 \).
2. It is rational when j is even and k is a multiple of 3.
3. For \( k=0 \): \( j=0,2,4,6 \) (4 terms); for \( k=3 \): \( j=0,2 \) (2 terms); for \( k=6 \): \( j=0 \) (1 term). Total 7.

Teacher's Note:
1. Use the multinomial expansion and list the allowed pairs (j, k).

Answer: (c) 101

Explanation:
1. \( T_{r+1}={}^{600}C_r\,7^{(600-r)/3}\,5^{r/2}x^r \).
2. The coefficient is an integer when \( 3\mid(600-r) \) and \( 2\mid r \), i.e. when r is a multiple of 6.
3. The values \( r=0,6,12,\dots,600 \) give \( 600/6+1=101 \) terms.

Teacher's Note:
1. Both exponents must be integers, so r has to be a multiple of 2 and of 3.

Answer: (c) 15

Explanation:
1. \( T_{r+1}={}^nC_r(-1)^rx^{3n-5r} \).
2. The coefficient of \( x^5 \) comes from \( 3n-5r=5 \) and that of \( x^{10} \) from \( 3n-5r'=10 \), so \( r'=r-1 \).
3. The condition \( {}^nC_r(-1)^r+{}^nC_{r-1}(-1)^{r-1}=0 \) needs \( {}^nC_r={}^nC_{r-1} \), i.e. \( n=2r-1 \). With \( 3n-5r=5 \): \( 6r-3-5r=5 \), \( r=8 \), \( n=15 \).

Teacher's Note:
1. Write the general term and find which r gives each power.
2. Equal binomial coefficients with adjacent r require n = 2r - 1.

Answer: (d) -30

Explanation:
1. Choose how many factors give \( x \) (say p) and how many give \( x^2 \) (say q) with \( p+2q=3 \): \( (p,q)=(3,0) \) or \( (1,1) \).
2. \( (3,0) \): \( \dfrac{5!}{3!\,2!}(-1)^3=-10 \). \( (1,1) \): \( \dfrac{5!}{1!\,1!\,3!}(-1)=-20 \).
3. The total is \( -10-20=-30 \).

Teacher's Note:
1. List the ways to reach the exponent 3 and use multinomial coefficients.

Answer: (a) \( {}^{16}C_9 \)

Explanation:
1. The series is a GP: \( \dfrac{(1+x)^{16}-(1+x)^6}{x} \).
2. The coefficient of \( x^6 \) is the coefficient of \( x^7 \) in \( (1+x)^{16}-(1+x)^6 \), which is \( {}^{16}C_7-0 \).
3. \( {}^{16}C_7={}^{16}C_9 \).

Teacher's Note:
1. Sum the GP in (1 + x) and shift the required power by one.

Answer: (a) 4

Explanation:
1. \( 1+x+x^2+x^3=\dfrac{1-x^4}{1-x} \), so the product is \( (1-x)^5\dfrac{(1-x^4)^4}{(1-x)^4}=(1-x)(1-x^4)^4 \).
2. \( (1-x^4)^4 \) has only powers \( x^{4k} \). For \( x^{13}=x\cdot x^{12} \) we need the term \( -x \) times the coefficient of \( x^{12} \) in \( (1-x^4)^4 \).
3. That coefficient is \( {}^4C_3(-1)^3=-4 \), so the answer is \( -1\times(-4)=4 \).

Teacher's Note:
1. Write the sum of the GP as (1 - x^4)/(1 - x) to cancel factors of (1 - x).

Answer: (c) 0

Explanation:
1. \( T_{r+1}={}^{12}C_r(-1)^r x^{2(12-r)-r}y^{-(12-r)+r}=\pm{}^{12}C_r\,x^{24-3r}y^{2r-12} \).
2. \( x^6 \) needs \( 24-3r=6 \), so \( r=6 \), which gives \( y^{0} \), not \( y^{-2} \).
3. The required term does not occur, so the coefficient is 0.

Teacher's Note:
1. Both exponents must match for the same value of r.

Answer: (c) \( -{}^{15}C_6 \)

Explanation:
1. \( T_{r+1}={}^{15}C_r(-1)^rx^{3(15-r)-2r}=\pm{}^{15}C_rx^{45-5r} \).
2. The constant term needs \( r=9 \).
3. Its value is \( {}^{15}C_9(-1)^9=-{}^{15}C_9=-{}^{15}C_6 \).

Teacher's Note:
1. Set the exponent of x equal to zero and use nCr = nC(n - r).

Answer: (a) \( {}^{22}C_{10} \)

Explanation:
1. \( \left(1+\dfrac1x\right)^{12}=\dfrac{(1+x)^{12}}{x^{12}} \), so the product is \( \dfrac{(1+x)^{22}}{x^{12}} \).
2. The constant term is the coefficient of \( x^{12} \) in \( (1+x)^{22} \), i.e. \( {}^{22}C_{12} \).
3. \( {}^{22}C_{12}={}^{22}C_{10} \).

Teacher's Note:
1. Combine the two binomials into a single power of (1 + x).

Answer: (a) \( {}^{11}C_5 \)

Explanation:
1. \( \left(x+\tfrac1x\right)^{10} \) contains only even powers \( x^{10-2r} \), so the term \( -2x \) of \( (1-x)^2=1-2x+x^2 \) contributes nothing to the constant term.
2. From 1: the constant term of the binomial is \( {}^{10}C_5 \). From \( x^2 \): the coefficient of \( x^{-2} \) is \( {}^{10}C_6={}^{10}C_4 \).
3. Sum \( {}^{10}C_5+{}^{10}C_4={}^{11}C_5 \).

Teacher's Note:
1. Match each term of the first factor with the matching power of x from the second.

Answer: (c) \( -\dfrac1{2^{13}}{}^{20}C_7x \)

Explanation:
1. The expansion has 21 terms, so the middle term is the 11th: \( r=11 \) and \( r+3=14 \).
2. \( T_{14}={}^{20}C_{13}(x^2)^{7}\left(-\tfrac1{2x}\right)^{13}=-\dfrac{{}^{20}C_{13}}{2^{13}}x^{14-13} \).
3. Since \( {}^{20}C_{13}={}^{20}C_7 \), \( T_{14}=-\dfrac1{2^{13}}{}^{20}C_7\,x \).

Teacher's Note:
1. Find r from the number of terms, then write the general term carefully with its sign.

Answer: (c) 197

Explanation:
1. \( (\sqrt2+1)^6+(\sqrt2-1)^6=2\left[{}^6C_0 2^3+{}^6C_2 2^2+{}^6C_4 2+{}^6C_6\right]=2(8+60+30+1)=198 \).
2. \( 0<(\sqrt2-1)^6<1 \), so \( (\sqrt2+1)^6=198-(\sqrt2-1)^6 \) lies between 197 and 198.
3. The greatest natural number below it is 197.

Teacher's Note:
1. Pair the number with its conjugate so the sum is an integer.

Answer: (a) \( \dfrac{1\cdot3\cdot5\cdots(2n-1)}{n!}\cdot2^n \)

Explanation:
1. The greatest coefficient is the middle one, \( {}^{2n}C_n=\dfrac{(2n)!}{n!\,n!} \).
2. Write \( (2n)!=[1\cdot3\cdot5\cdots(2n-1)]\cdot[2\cdot4\cdots2n]=[1\cdot3\cdots(2n-1)]\,2^nn! \).
3. So \( {}^{2n}C_n=\dfrac{1\cdot3\cdot5\cdots(2n-1)}{n!}2^n \).

Teacher's Note:
1. Split (2n)! into the product of odd and even numbers.

Answer: (c) 2

Explanation:
1. \( 99^n+1=(100-1)^n+1=100n+{}^nC_2100^2-\dots \) because n is odd, so the terms \( -1+1 \) cancel.
2. Every term after \( 100n \) is a multiple of \( 100^2 \).
3. So \( 99^n+1\equiv100n \pmod{10^4} \); for odd n, n is not a multiple of 10, so the number ends in exactly two zeros.

Teacher's Note:
1. Write 99 as 100 - 1 and keep only the lowest-order surviving terms.

Answer: (a) \( 2^{18} \)

Explanation:
1. The 20 coefficients are \( {}^{19}C_0,\dots,{}^{19}C_{19} \), and the last ten are \( {}^{19}C_{10},\dots,{}^{19}C_{19} \).
2. By symmetry \( {}^{19}C_r={}^{19}C_{19-r} \), these equal the first ten, \( {}^{19}C_0,\dots,{}^{19}C_9 \).
3. So each half sums to \( \tfrac12\cdot2^{19}=2^{18} \).

Teacher's Note:
1. Use the symmetry of the binomial coefficients about the middle.

Answer: (d) \( \tfrac12(3^{40}+1) \)

Explanation:
1. \( T_{r+1}={}^{40}C_r2^rx^{r/2} \) has an integral power of x when r is even.
2. The sum of the even-r terms is \( \tfrac12\left[(1+2)^{40}+(1-2)^{40}\right] \).
3. That is \( \tfrac12(3^{40}+1) \).

Teacher's Note:
1. Sum of even-indexed terms equals half of f(1) + f(-1) in terms of the variable 2.

Answer: (d) none of these

Explanation:
1. By symmetry \( {}^{20}C_r={}^{20}C_{20-r} \), so \( 2\sum_{r=0}^{9}{}^{20}C_r+{}^{20}C_{10}=2^{20} \).
2. Then \( \sum_{r=0}^{10}{}^{20}C_r=\sum_{r=0}^{9}{}^{20}C_r+{}^{20}C_{10}=2^{19}-\tfrac12{}^{20}C_{10}+{}^{20}C_{10} \).
3. This equals \( 2^{19}+\tfrac12{}^{20}C_{10}=2^{19}+\tfrac12\cdot\dfrac{20!}{(10!)^2} \), which matches none of the options.

Teacher's Note:
1. Use the symmetry nCr = nC(n - r) to split the full sum into two equal halves plus the middle term.

Answer: (c) \( {}^{21}C_{17}-{}^{10}C_6 \)

Explanation:
1. The hockey-stick identity gives \( \sum_{k=3}^{20}{}^{k}C_3={}^{21}C_4 \).
2. Remove the terms from \( k=3 \) to \( 9 \): \( \sum_{k=3}^{9}{}^kC_3={}^{10}C_4 \).
3. So the sum is \( {}^{21}C_4-{}^{10}C_4={}^{21}C_{17}-{}^{10}C_6 \).

Teacher's Note:
1. Apply the identity that the sum of kC3 from k = 3 to n equals (n + 1)C4 and subtract the unwanted part.

Answer: (a) \( \tfrac12 \)

Explanation:
1. The terms are \( \tfrac12(-2)^k\,{}^{10}C_k \): for k = 0 it is \( \tfrac12 \), for k = 1 it is \( -{}^{10}C_1 \), and for k = 10 it is \( \tfrac12\cdot2^{10}=2^9 \).
2. The sum is \( \tfrac12\sum_{k=0}^{10}{}^{10}C_k(-2)^k=\tfrac12(1-2)^{10} \).
3. Therefore the value is \( \tfrac12 \).

Teacher's Note:
1. Spot that the general term is half of nCk times (-2) to the power k.

Answer: (c) \( n\cdot2^{n-1} \)

Explanation:
1. Use \( r\,{}^nC_r=n\,{}^{n-1}C_{r-1} \).
2. The sum becomes \( n\sum_{r=1}^n{}^{n-1}C_{r-1}=n\cdot2^{n-1} \).

Teacher's Note:
1. Use the identity r nCr = n (n - 1)C(r - 1).

Answer: (a) \( n\cdot2^{2n-1} \)

Explanation:
1. \( r\,{}^{2n}C_r=2n\,{}^{2n-1}C_{r-1} \), so the sum is \( 2n\sum_{j=0}^{n-1}{}^{2n-1}C_j \).
2. The coefficients \( {}^{2n-1}C_0,\dots,{}^{2n-1}C_{n-1} \) are exactly half of the \( 2n \) coefficients of \( (1+x)^{2n-1} \), so they sum to \( 2^{2n-2} \).
3. The result is \( 2n\cdot2^{2n-2}=n\cdot2^{2n-1} \).

Teacher's Note:
1. Use the symmetry of the coefficients of (1 + x) raised to an odd power 2n - 1.

Answer: (b) 0

Explanation:
1. \( r\,{}^{20}C_r=20\,{}^{19}C_{r-1} \), so the sum is \( 20\sum_{r=1}^{20}(-1)^{r-1}{}^{19}C_{r-1} \).
2. That is \( 20\,(1-1)^{19}=0 \).
3. Hence the value is 0.

Teacher's Note:
1. Reduce r nCr to n times (n - 1)C(r - 1) and use the alternating sum of binomial coefficients.