JEE Mathematics Binomial Theorem MCQs Set 11

Answer: (c) \({}^{31}C_6-{}^{21}C_6\)

Explanation:
1. The coefficient is \(\sum_{m=21}^{30}{}^mC_5=\sum_{m=5}^{30}{}^mC_5-\sum_{m=5}^{20}{}^mC_5\).
2. By the hockey-stick identity \(\sum_{m=5}^{N}{}^mC_5={}^{N+1}C_6\), so the result is \({}^{31}C_6-{}^{21}C_6\).

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Answer: (d) None of these

Explanation:
1. Expand \((x-r)(y-r)=xy-(x+y)r+r^2\). The sum is \(xy\,S_0-(x+y)S_1+S_2\) with \(S_k=\sum(-1)^r\,{}^nC_r\,r^k\).
2. For \(n>2\), \(S_0=S_1=S_2=0\) (they are the values of \((1-t)^n\) and its first two derivative-type combinations at \(t=1\)), so the total is \(0\), which matches none of (a), (b), (c).

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Answer: (a) 0

Explanation:
1. Since \({}^nC_r={}^nC_{n-r}\), the terms \(r\) and \(n-r\) have equal magnitude.
2. For odd \(n\), \((-1)^{n-r}=-(-1)^r\), so they cancel in pairs and the sum is \(0\).

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Answer: (c) \(\dfrac1{n+1}\), \(n\) is even

Explanation:
1. \(\dfrac{{}^nC_r}{{}^{r+2}C_r}=\dfrac{2\,{}^nC_r}{(r+1)(r+2)}\) and \(\dfrac{{}^nC_r}{(r+1)(r+2)}=\dfrac{{}^{n+2}C_{r+2}}{(n+1)(n+2)}\).
2. The sum becomes \(\dfrac{2}{(n+1)(n+2)}\sum_{r=0}^n(-2)^r\,{}^{n+2}C_{r+2}=\dfrac{2}{4(n+1)(n+2)}\sum_{k=2}^{n+2}(-2)^k\,{}^{n+2}C_k\).
3. \(\sum_{k=2}^{n+2}(-2)^k\,{}^{n+2}C_k=(1-2)^{n+2}-1+2(n+2)=(-1)^{n+2}+2n+3\). For even \(n\) this is \(2n+4\), giving \(\dfrac{2(2n+4)}{4(n+1)(n+2)}=\dfrac1{n+1}\); for odd \(n\) it is \(2n+2\), giving \(\dfrac1{n+2}\).
4. So the value is \(\dfrac1{n+1}\) when \(n\) is even: option (c).

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Answer: (c) 1

Explanation:
1. Since \({}^{57}C_{53}={}^{57}C_4\), put the three groups over \({}^{57}C_4\). Also \({}^{56-j}C_{53-j}={}^{56-j}C_3\).
2. The numerator becomes \({}^{47}C_4+\sum_{m=47}^{50}{}^mC_3+\sum_{m=51}^{56}{}^mC_3={}^{47}C_4+\sum_{m=47}^{56}{}^mC_3\).
3. By the hockey-stick identity \({}^{47}C_4+\sum_{m=47}^{56}{}^mC_3={}^{57}C_4\), so the value is \(1\).

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Answer: (a) \(ab=1\)

Explanation:
1. First expansion: \(T_{r+1}={}^8C_r\,a^{8-r}b^{-r}x^{16-3r}\). For \(x^7\): \(r=3\), coefficient \(56\,a^5b^{-3}\).
2. Second: \(T_{r+1}={}^8C_r\,a^{8-r}(-1)^rb^{-r}x^{8-3r}\). For \(x^{-7}\): \(r=5\), magnitude \(56\,a^3b^{-5}\).
3. \(a^5b^{-3}=a^3b^{-5}\Rightarrow a^2b^2=1\Rightarrow ab=1\).

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Answer: (c) 17

Explanation:
1. For \(k\ge5\), \(k!\) is divisible by \(10\) (it contains the factors 2 and 5), so \((k!)^2\) is divisible by \(100\).
2. Only \(k=1,2,3,4\) matter: \(1+4+36+576=617\equiv17\pmod{100}\).

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Answer: (c) 120

Explanation:
1. \(\dfrac{a_r}{a_{r-1}}=\dfrac{16-r}{r}\), so \(r\dfrac{a_r}{a_{r-1}}=16-r\).
2. \(\sum_{r=1}^{15}(16-r)=15+14+\cdots+1=120\).

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Answer: (b) \({}^nC_r\)

Explanation:
1. \({}^{n-2}C_r+{}^{n-2}C_{r-1}={}^{n-1}C_r\) and \({}^{n-2}C_{r-1}+{}^{n-2}C_{r-2}={}^{n-1}C_{r-1}\).
2. Adding, \({}^{n-1}C_r+{}^{n-1}C_{r-1}={}^nC_r\).

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Answer: (b) 0

Explanation:
1. \({}^{15}C_8={}^{15}C_7\) and \({}^{15}C_9={}^{15}C_6\), so the expression is \(0\).

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Answer: (a) \(\dfrac{2017!}{17!\,2000!}\)

Explanation:
1. The terms form a G.P. with first term \((1+x)^{2016}\), ratio \(\dfrac x{1+x}\) and \(2017\) terms. Its sum is \((1+x)^{2017}-x^{2017}\).
2. The coefficient of \(x^{17}\) is \({}^{2017}C_{17}=\dfrac{2017!}{17!\,2000!}\).

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Answer: (b) \(7^{th}\)

Explanation:
1. Let the terms be \(T_{r+1},T_{r+2},T_{r+3}\) with coefficients \({}^nC_r,{}^nC_{r+1},{}^nC_{r+2}\).
2. \(\dfrac{n-r}{r+1}=7\) and \(\dfrac{n-r-1}{r+2}=6\). Then \(n-r=7r+7\) and \(n-r-1=6r+12\), giving \(r=6,\ n=55\).
3. The first term is \(T_7\), the 7th.

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Answer: (b) \(8^{th}\) and \(9^{th}\)

Explanation:
1. Coefficient of \(x^r\) is \({}^{55}C_r2^{55-r}3^{-r}\). Equality for \(r\) and \(r+1\) gives \(\dfrac{55-r}{r+1}\cdot\dfrac1{6}=1\).
2. \(55-r=6r+6\Rightarrow r=7\). The terms are those with \(x^7\) and \(x^8\), the \(8^{th}\) and \(9^{th}\).

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Answer: (d) 1 : 32

Explanation:
1. \(T_{r+1}={}^{15}C_r\,2^r\,x^{30-3r}\). For \(x^{15}\): \(r=5\); for the constant term: \(r=10\).
2. Ratio \(=\dfrac{{}^{15}C_5\,2^5}{{}^{15}C_{10}\,2^{10}}=\dfrac1{32}\), since \({}^{15}C_5={}^{15}C_{10}\).

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Answer: (a) 182

Explanation:
1. \(T_{r+1}={}^{18}C_r\,2^{-r}x^{(18-2r)/3}\). For \(x^{-2}\): \(r=12\); for \(x^{-4}\): \(r=15\).
2. \(\dfrac mn=\dfrac{{}^{18}C_{12}\,2^{-12}}{{}^{18}C_{15}\,2^{-15}}=\dfrac{18564}{816}\times8=182\).

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Answer: (a) 400

Explanation:
1. \(\left(2x^2-\dfrac1x\right)^8\) has general term \({}^8C_r2^{8-r}(-1)^rx^{16-3r}\).
2. Constant term needs: \(1\cdot x^0\) (\(16-3r=0\), none); \(-\dfrac1x\cdot x^1\) (\(16-3r=1\Rightarrow r=5\)): \(-{}^8C_5\,2^3(-1)^5=448\); \(3x^5\cdot x^{-5}\) (\(16-3r=-5\Rightarrow r=7\)): \(3\cdot{}^8C_7\,2\,(-1)^7=-48\).
3. Total \(=448-48=400\).

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Answer: (b) 6

Explanation:
1. Coefficients: \({}^{18}C_{2r+3}={}^{18}C_{r-3}\). Either \(2r+3=r-3\) (impossible, \(r=-6\)) or \((2r+3)+(r-3)=18\), so \(r=6\).

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Answer: (c) 60

Explanation:
1. \(x^5\) arises from \(x^0\cdot x^5\) (impossible, degree 4), \(x^2\cdot x^3\), \(x^4\cdot x^1\): \(5\cdot4+10\cdot4=20+40=60\).

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Answer: (a) 9

Explanation:
1. With \((1+x^2)^2=1+2x^2+x^4\): \(a_1={}^nC_1=n\), \(a_2={}^nC_2+2\), \(a_3={}^nC_3+2\,{}^nC_1\).
2. \(2a_2=a_1+a_3\) gives \(n(n-1)+4=n+\dfrac{n(n-1)(n-2)}6+2n\). Multiply by 6 and simplify: \(n^3-9n^2+26n-24=0\), i.e. \((n-2)(n-3)(n-4)=0\).
3. The possible values are \(2,3,4\), with sum \(9\).

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Answer: (b) 1

Explanation:
1. Let \(R'=(7-4\sqrt3)^n\in(0,1)\). \(R+R'\) is an integer, so \(\beta+R'=1\), i.e. \(1-\beta=R'\).
2. \((1-\beta)(p+\beta)=R'R=\big[(7-4\sqrt3)(7+4\sqrt3)\big]^n=1\).

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Answer: (c) 243

Explanation:
1. \((a-1)^n=\sum_{r=0}^n(-1)^r\,{}^nC_ra^{n-r}=a\,f(n)+(-1)^n\), so \(f(n)=\dfrac{(a-1)^n-(-1)^n}a\).
2. \(f(2007)+f(2008)=\dfrac{3^{2007/223}+1+3^{2008/223}-1}a=\dfrac{3^9(1+3^{1/223})}a=3^9=19683\).
3. \(81k=19683\Rightarrow k=243\).

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61. b
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65. .

Answer: (a) 40

Explanation:
1. The sum is \(\sum_{r}(-1)^r\binom{30}r\binom{30}{20-r}\), the coefficient of \(x^{20}\) in \((1-x)^{30}(1+x)^{30}=(1-x^2)^{30}\).
2. That coefficient is \((-1)^{10}\,{}^{30}C_{10}={}^{30}C_{10}\). With \(y\le10\): \(x=30,y=10\), so \(x+y=40\).

Teacher's Note:
1. U
2. s
3. e
4.
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6. 1
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10. x
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12. ^
13. 3
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31.
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33.
34. x
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38. ^
39. 3
40. 0
41. .

Answer: (b) 98

Explanation:
1. \(2\,{}^nC_5={}^nC_4+{}^nC_6\) leads to \(n^2-21n+98=0\), so \(n=7\) or \(14\).
2. Product \(=98\).

Teacher's Note:
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47. .

Answer: (c) 17

Explanation:
1. \(T_{r+1}={}^{100}C_rx^{(100-r)/2}y^{r/3}\) needs \(r\) divisible by \(3\) and \(100-r\) even, i.e. \(r\) a multiple of \(6\).
2. \(r=0,6,\dots,96\): \(17\) terms.

Teacher's Note:
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31. e
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35. .

Answer: (a) 6

Explanation:
1. \(na=8\) and \({}^nC_2a^2=24\). Dividing: \(\dfrac{n-1}{2n}\cdot8=3\), so \(n=4\), \(a=2\).
2. \(a+n=6\).

Teacher's Note:
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Answer: (d) 10

Explanation:
1. \(1+2x+3x^2+\cdots=(1-x)^{-2}\), so the square is \((1-x)^{-4}\).
2. The coefficient of \(x^2\) is \({}^{4+2-1}C_2={}^5C_2=10\).

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27. .

Answer: (a) 2

Explanation:
1. Middle term of \((1+kx)^6\) is the 4th: \({}^6C_3k^3=20k^3\). Middle term of \((1-kx)^8\) is the 5th: \({}^8C_4k^4=70k^4\).
2. \(20k^3=70k^4\Rightarrow k=\dfrac27\), so \(7k=2\).

Teacher's Note:
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37. .

Answer: (c) 7

Explanation:
1. \(T_{r+1}={}^8C_r\left(\tfrac12\right)^{8-r}x^{(8-r)/3-r/5}\). The exponent is \(0\) when \(5(8-r)=3r\), so \(r=5\).
2. \(T_6={}^8C_5\left(\tfrac12\right)^3=\dfrac{56}8=7\).

Teacher's Note:
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