JEE Mathematics Binomial Theorem MCQs Set 08

Answer: (d) 61

Explanation:
1. \( 1-3x+3x^2-x^3=(1-x)^3 \)
2. So the expression is \( [(1-x)^3]^{20}=(1-x)^{60} \)
3. It has \( 60+1=61 \) terms

Teacher's Note:
1. Recognise the cube of a binomial before expanding.
2. A binomial power n has n + 1 terms.

Answer: (c) \( \frac{10500}{x^3} \)

Explanation:
1. \( T_7={}^9C_6\left(\frac{4x}5\right)^3\left(-\frac5{2x}\right)^6 \)
2. \( =84\cdot\frac{64x^3}{125}\cdot\frac{15625}{64x^6} \)
3. \( =84\cdot125\,x^{-3}=\frac{10500}{x^3} \) (the sixth power makes the sign positive)

Teacher's Note:
1. Keep the minus sign inside the bracket before raising to a power.
2. Even powers of a negative term are positive.

Answer: (b) 501

Explanation:
1. \( T_{r+1}={}^{1000}C_r\,3^{(1000-r)/2}\,2^{r/2} \)
2. Both exponents must be integers, so r is even
3. \( r=0,2,4,\dots,1000 \) gives 501 values

Teacher's Note:
1. Write each radical as a power of a prime.
2. Count the values of r that make all exponents integers.

Answer: (a) \( \frac{189}8a^{17} \) and \( -\frac{21}{16}a^{19} \)

Explanation:
1. \( n=9 \) is odd, so the middle terms are \( T_5 \) and \( T_6 \)
2. \( T_5={}^9C_4(3a)^5\left(-\frac{a^3}6\right)^4=126\cdot243\cdot\frac1{1296}a^{17}=\frac{189}8a^{17} \)
3. \( T_6={}^9C_5(3a)^4\left(-\frac{a^3}6\right)^5=-126\cdot81\cdot\frac1{7776}a^{19}=-\frac{21}{16}a^{19} \)

Teacher's Note:
1. For odd n there are two middle terms, the \( \frac{n+1}2 \)th and the next.
2. Evaluate each separately.

Answer: (a) \( T_4 \) and \( T_5 \), each of magnitude \( {}^{15}C_3\,3^{12} \)

Explanation:
1. \( \left
2. \frac{T_{r+1}}{T_r}\right
3. =\frac{16-r}{r}\cdot\frac{
4. 5x
5. }{3}=\frac{16-r}{3r} \) at \( x=\frac15 \)
6. This is \( \ge1 \) when \( r\le4 \), and equals 1 at \( r=4 \)
7. So \( T_5=T_4 \) in magnitude, and they are the greatest
8. Magnitude \( ={}^{15}C_3\,3^{12}=241805655 \)

Teacher's Note:
1. Compare consecutive terms through their ratio.
2. When the ratio equals 1 at some r, two terms are equally greatest.

Answer: (c) 77587665

Explanation:
1. The terms with \( r_2 \) powers of x and \( r_3 \) powers of \( \frac7x \) need \( r_2=r_3 \)
2. The term is \( 1+\sum_{r=1}^{5}{}^{11}C_{2r}\,{}^{2r}C_r\,7^r \)
3. Evaluating the sum gives 77587665

Teacher's Note:
1. A multinomial expansion splits n among the three parts.
2. Equal powers of x and 1/x give a constant.

Answer: (a) \( 1-\frac{41}{24}x \)

Explanation:
1. Numerator \( \approx1-\frac32x+1-\frac53x=2-\frac{19}6x \)
2. Denominator \( =2\left(1+\frac x4\right)^{1/2}\approx2\left(1+\frac x8\right) \)
3. Ratio \( \approx\frac12\left(2-\frac{19}6x\right)\left(1-\frac x8\right)=1-\frac x8-\frac{19}{12}x=1-\frac{41}{24}x \)

Teacher's Note:
1. Expand each power to the first order in x.
2. Multiply and drop terms with x squared.

Answer: (a) 001

Explanation:
1. \( 81^{25}=(1+80)^{25}=1+25\cdot80+{}^{25}C_2\cdot80^2+\dots \)
2. \( 25\cdot80=2000 \) and \( {}^{25}C_2\cdot6400=1920000 \), and the rest are multiples of \( 80^3 \)
3. All of these are multiples of 1000
4. So the last three digits are 001

Teacher's Note:
1. Write 81 as 1 + 80 and expand.
2. Only the first few terms affect the last three digits.

Answer: (c) \( (1.2)^{4000} \)

Explanation:
1. \( (1.2)^{4000}=(1+0.2)^{4000}>1+4000(0.2)=801 \), because all binomial terms are positive
2. \( 801>800 \)

Teacher's Note:
1. Keep only the first two terms of the expansion as a lower bound.
2. This works when all terms are positive.

Answer: (c) 4

Explanation:
1. The series \( 1+t+\frac{1\cdot3}{2!}t^2+\frac{1\cdot3\cdot5}{3!}t^3+\dots \) is \( (1-2t)^{-1/2} \)
2. With \( t=\frac25 \): \( 1+y=\left(1-\frac45\right)^{-1/2}=\sqrt5 \)
3. \( y^2+2y=(1+y)^2-1=5-1=4 \)

Teacher's Note:
1. Match the series with the expansion of (1 - 2t) to the power minus one half.
2. Complete the square to reach y squared plus 2y.

Answer: (c) -197

Explanation:
1. \( \frac1{(1-x)^2}=\sum(k+1)x^k \)
2. Coefficient of \( x^{100} \) \( =3(101)-5(100)=303-500=-197 \)

Teacher's Note:
1. Expand 1 over (1 - x) squared first.
2. Collect the two contributions to x to the hundredth.

Answer: (c) 12

Explanation:
1. Coefficient of x: \( m-n=3 \)
2. Coefficient of \( x^2 \): \( {}^mC_2+{}^nC_2-mn=-6 \)
3. Put \( n=m-3 \): \( \frac{m(m-1)}2+\frac{(m-3)(m-4)}2-m(m-3)=-6 \), which gives \( m=12 \)

Teacher's Note:
1. Write both coefficients as equations in m and n.
2. Substitute n = m - 3 to get a linear equation.

Answer: (b) 5

Explanation:
1. Odd powers of x cancel in the sum of the two expansions
2. Only the even powers \( x^0,x^2,x^4,x^6,x^8 \) remain
3. So there are 5 terms

Teacher's Note:
1. The two expansions differ only in the sign of odd-power terms.
2. Even-power terms double.

Answer: (c) \( \log_e\frac4e \)

Explanation:
1. \( \frac1{k(k+1)}=\frac1k-\frac1{k+1} \), so the sum is \( \left(1-\frac12\right)-\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)-\dots \)
2. \( =1-2\left(\frac12-\frac13+\frac14-\dots\right) \)
3. \( \frac12-\frac13+\frac14-\dots=1-\log_e2 \), so the sum is \( 1-2(1-\log_e2)=2\log_e2-1=\log_e\frac4e \)

Teacher's Note:
1. Split into partial fractions and regroup.
2. Use the series for log 2.

Answer: (a) \( \frac12\left(e+\frac1e\right) \)

Explanation:
1. \( e=1+\frac1{1!}+\frac1{2!}+\dots \) and \( e^{-1}=1-\frac1{1!}+\frac1{2!}-\dots \)
2. Adding: \( e+e^{-1}=2\left(1+\frac1{2!}+\frac1{4!}+\dots\right) \)
3. The sum is \( \frac12(e+e^{-1}) \)

Teacher's Note:
1. Add the series for e and e to the minus 1 so that odd terms cancel.
2. The result is the hyperbolic cosine of 1.

Answer: (a) 5 or 9

Explanation:
1. The coefficients are \( {}^{14}C_{r-1},{}^{14}C_r,{}^{14}C_{r+1} \)
2. A.P.: \( 2\,{}^{14}C_r={}^{14}C_{r-1}+{}^{14}C_{r+1} \)
3. This simplifies to \( r^2-14r+45=0 \)
4. So \( r=5 \) or \( 9 \)

Teacher's Note:
1. Convert coefficients to factorial form and simplify with the ratio formula.
2. Solve the resulting quadratic.

Answer: (b) 55

Explanation:
1. Let the coefficients be \( {}^nC_{r-1},{}^nC_r,{}^nC_{r+1} \)
2. \( \frac{{}^nC_r}{{}^nC_{r-1}}=\frac{n-r+1}r=7 \) and \( \frac{{}^nC_{r+1}}{{}^nC_r}=\frac{n-r}{r+1}=6 \)
3. So \( n+1=8r \) and \( n=7r+6 \), giving \( r=5 \), \( n=55 \)

Teacher's Note:
1. Use the ratio of consecutive binomial coefficients.
2. Two equations in n and r give both.

Answer: (c) e-2

Explanation:
1. \( \sum_{n=1}^\infty\frac1{(n+1)!}=\frac1{2!}+\frac1{3!}+\dots \)
2. \( e=1+\frac1{1!}+\frac1{2!}+\dots \) so the sum is \( e-1-1=e-2 \)

Teacher's Note:
1. Compare with the series for e and subtract the missing terms.
2. Two terms are missing, 1 and 1.

Answer: (a) \( \frac12\left(e-\frac1e\right) \)

Explanation:
1. \( e=\sum\frac1{k!} \) and \( e^{-1}=\sum\frac{(-1)^k}{k!} \)
2. Subtracting: \( e-e^{-1}=2\left(\frac1{1!}+\frac1{3!}+\frac1{5!}+\dots\right) \)
3. The sum is \( \frac12\left(e-\frac1e\right) \)

Teacher's Note:
1. Subtract the two series so that even terms cancel.
2. The result is the hyperbolic sine of 1.

Answer: (c) 4th

Explanation:
1. \( T_{r+1}={}^5C_r(x^2)^{5-r}x^{-r}=\binom5rx^{10-3r} \)
2. \( 10-3r=1\Rightarrow r=3 \)
3. It is the 4th term

Teacher's Note:
1. Collect the power of x in the general term.
2. The term number is r + 1.

Answer: (c) 1001

Explanation:
1. \( T_{r+1}={}^{15}C_r(-1)^rx^{\frac{15-r}3-\frac r2} \)
2. Constant when \( \frac{15-r}3=\frac r2\Rightarrow r=6 \)
3. The term is \( {}^{15}C_6=5005=5\times1001 \), so \( m=1001 \)

Teacher's Note:
1. Set the exponent of x to zero.
2. Check the sign (-1) to the sixth power is positive.