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BINOMIAL THEOREM - Studiestoday...Find the middle terms in the expansion of @3ð¥â 3 6 A? Mcqs e)...
Transcript of BINOMIAL THEOREM - Studiestoday...Find the middle terms in the expansion of @3ð¥â 3 6 A? Mcqs e)...
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BINOMIAL THEOREM
QUESTIONS BASED ON âGENERAL TERMSâ
Q.1) Find the positive value of âmâ for which the coefficient of ð¥2 in the expansion of (1 + ð¥)ð is 6?
mcqs a) 6 b) 9 c) 4 d) 1
Sol.1) Given expansion: (1 + ð¥)ð coefficient of ð¥2 = 6 To find : m General term: ðð+1= mðð(1)ðâðð¥ð = ðð+1= mððð¥ð For ð¥2 put r=2 ⎠ð3 = mð2ð¥ð Here, coefficient of ð¥2 = 6 â mð2 = 6
â ð(ð â 1)
2= 6
â ð2-m-12=0 â (ð â 4)(ð + 3) = 0 â m=4 or m= -3 but m cannot negative(-) ⎠m â -3 ⎠m= 4 ans.
Q.2) If the coefficient of (ð â 5)ð¡â and (2ð â 1)ð¡â terms in this expansion of (1 + ð¥)34 are equal. Find the value of ârâ?
mcqs a) 14 b) 10 c) 12 d) 20
Sol.2) Given expansion: (1 + ð¥)34 Coefficient of ððâ5 = ð2ðâ1 To find ârâ General term: ðð+1 = 34ððð¥2(1)34âðð¥ð = ðð+1 = 34ððð¥2 For ððâ5 , put r = r-6 ⎠ððâ5 = 34ððâ6 ð¥
ðâ6 Here coefficient of ððâ5 = 34ððâ6 For ð2ðâ1 , put r = 2r-2 ⎠ð2ðâ2 = 34ð2ðâ2 ð¥
2ðâ2 Here coefficient of ð2ðâ1 = 34ð2ðâ2 We are given that, coefficients are equal â 34ððâ6 = 34ð2ðâ2 â r-6 + 2r-2 = 34âŠâŠâŠ..(if nðð¥ = nððŠ then x+y=n or x=y)
(Or) r-6 = 2r-2 â 3r=42 or r â 4 but ârâ cannot be negative(-) â r=14 ans.
Q.3) Find the term independent of x in the expansion of (
3ð¥2
2â
1
3ð¥)
6
?
mcqs
a) â9
1 b)
5
12 c) 10 d)
2
8
Sol.3) General term is given by ðð+1 = (â1)ð . 6ðð (3)6â2ð
26âð . ð¥12â3ð
For independent term of x i.e. ð¥0 , put 12-3r = 0 â r = 4
⎠ð5 = (â1)4 . 6ð4 (3)6â8
22 . ð¥0
= 6ð2 (3)â2
22 âŠâŠâŠâŠ (6ð4 = 6ð2 )
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m
= 6 Ã5
2 Ã
1
9 Ã4 =
5
12
⎠5th term is the independent term of x and is given by 5
12 ans.
Q.4) Find the value of âaâ so that the term independent of âxâ in (âð¥ +
ð
ð¥2)
10 is 405?
Mcqs a) ð2 = 40Ã9
8Ã11 b) ð2 =
â405Ã2
9Ã10
c) ð2 = 405Ã2
9Ã10 d) ð2 =
205Ã2
3Ã6
Sol.4) Given expansion: (ð¥
12â +
ð
ð¥2)10
Independent term of x = 405 To find âaâ
General term: ðð+1 = 10ðð (ð¥1
2â )10âð
. ðð
ð¥2ð
â ðð+1 = 10ðð (ð¥)10âð
2â2ð.
ðð
ð¥2ð
â ðð+1 = 10ðð (ð¥)10âð
2â2ð. ðð
â ðð+1 = 10ðð (ð¥)10â5ð
2 . ðð
Now, for independent term of x i.e. ð¥0 , Put 10â5ð
2= 0
â r = 2 ⎠ð3 = 10ð2 (ð¥)0. ð2 ð3 = 10ð2 ð
2 Also independent term of x = 405âŠâŠâŠâŠ(given) â 10ð2 ð
2 = 405
â 10Ã9
2 ð2 = 405
â ð2 = 405Ã2
9Ã10 ans.
Q.5) Find the middle terms in the expansion of (3ð¥ â
ð¥3
6)
7
?
Mcqs e) 42ð¥13 and 35 f)
â105
8ð¥13 and
35
48ð¥15
g) 25
72ð¥13 and
30
48 h)
â10
1ð¥1 and
35
8ð¥5
Sol.5) Given expansion: (3x â
x3
6)
7
To find âmiddle termâ
Since, power is odd, ⎠there are two middle terms = (n+1
2)
th and (
n+3
2)
th
i.e. (7+1
2)
ð¡â and (
7+3
2)
ð¡â
â 4th and 5th terms
General term: ðð+1 = (â1)ð 7ðð (3ð¥)7âð (ð¥3
6)
2
= (â1)ð 7ðð (3)7âð . ð¥7âð . ð¥3ð
6ð
= ðð+1 = (â1)ð 7ðð (3)7âð
6ð ð¥7+2ð
For ð4 , put r = 3
= T3 = (â1)ð 7ð3 (3)4
63 ð¥7+6
= â7Ã6Ã5
6 Ã
81
216 ð¥13
= T4= â105
8ð¥13
For ð5 , put r = 4
⎠ð5 = (â1)ð 7ð4 (3)3
64 ð¥7+8
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m
= ð5 = 35
48ð¥15
⎠the middle terms are â105
8ð¥13 and
35
48ð¥15 ans.
Q.6) Show that the middle term in the expansion of (1 + ð¥)2ð is 1.3.5âŠâŠâŠ(2ðâ1).2ð.ð¥ð
ð!?
Mcqs a) False b) True c) Not proved d) Negative
Sol.6) Given expansion (1 + ð¥)2ð
Since, power (2ð) is even, only 1 middle term = (2ð
2+ 1)
ð¡â = (ð + 1)ð¡â terms
General term: ðð+1 = 2nððð¥ð For ðð+1 , put r = n = ðð+1 = 2nððð¥ð
= (2ð)!
ð!ð!ð¥ð
= 1.2.3.4.5.6âŠâŠâŠ(2ðâ1).(2ð).ð¥ð
ð!ð!
= [1.3.5âŠâŠâŠ(2ðâ1)] [2.4.6âŠâŠâŠ(2ð)ð¥ð]
ð!ð!
= [1.3.5âŠâŠâŠ(2ðâ1)].2ð (1.2.3âŠâŠâŠð).ð¥ð
ð!ð!
= 1.3.5âŠâŠâŠ(2ðâ1).2ð.ð!.ð¥ð
ð!ð!
= ðð+1 = 1.3.5âŠâŠâŠ(2ðâ1).2ð.ð¥ð
ð! ans.
Q.7) Show that the coefficient of the middle terms in the expansion of (1 + ð¥)2ð is equal to the sum of the coefficients of two middle terms in the expansion of (1 + ð¥)2ðâ1? For mcqsâŠ.. is it true or false?
Mcqs e) True f) False g) Negative h) Positive
Sol.7) 1st expansion : (1 + ð¥)2ð Since, power (2ð) is even , only 1 middle term
= (2ð
2+ 1)
ð¡â = (ð + 1)ð¡â term
General term: ðð+1 = 2nððð¥ð For ðð+1 , put r = n = ðð+1 = 2nððð¥ð..............(coefficient = 2nðð) 2nd expansion: (1 + ð¥)2ðâ1 Since, power (2ð â 1) is odd , only 2 middle terms
= (2ðâ1+1
2)
ð¡â = (
2ðâ1+3
2)
ð¡â term
= ðð¡âand (ð + 1)ð¡â terms General term: ðð+1 = 2n-1ððð¥ð For ðð , put r = n-1 ⎠ðð= 2n-1ððâ1 ð¥
ðâ1 Coefficient = 2n-1ððâ1 For ðð+1 , put r = n ⎠ðð+1 = 2n-1ðð ð¥
ð Coefficient = 2n-1ðð Now, we have to prove that 2nðð = 2n-1ððâ1 + 2n-1ðð R.H.S = 2n-1ððâ1 + 2n-1ðð = 2n-1+1ðð âŠâŠâŠ(nðð+ nððâ1= n+1ðð) = 2nðð = L.H.S (proved)
Q.8) Prove that the coefficient of ð¥ð is the expansion of (1 + ð¥)2ð is twice the coefficient of ð¥ð in the expansion of(1 + ð¥)2ðâ1?
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For mcqsâŠ.. is it true or false?
Mcqs i) True j) False k) Negative l) Positive
Sol.8) 1st expansion: (1 + ð¥)2ð General term: ðð+1 = 2nððð¥ð For ð¥ð , put r = n = ðð+1 = 2nððð¥ð Coefficient of ð¥ð = 2nðð 2nd expansion: (1 + ð¥)2ðâ1 General term: ðð+1 = 2n-1ððð¥ð For ð¥ð , put r = n = ðð+1 = 2n-1ððð¥ð Coefficient of ð¥ð = 2n-1ðð Now, we have to prove that = 2nðð = 2(2n-1ðð) R.H.S = 2. 2n-1ðð
= 2(2ðâ1)!
ð!(ðâ1)! âŠâŠâŠâŠ..(1)
L.H.S = 2nðð
= (2ð)!
ð!ð! =
(2ð)(2ðâ1)!
ð!ð(ðâ1)!
= 2(2ðâ1)!
ð!(ðâ1)! âŠâŠâŠâŠâŠ..(2)
From (1) & (2) , R.H.S = L.H.S (proved)
Q.9) The sum of the coefficients of the 1st three terms in the expansion of (ð¥ â3
ð¥2)ð
is 559.
Find the term containing ð¥3 in the expansion?
mcqs a) 2582ð¥3 b) -5940ð¥3 c) 5900 d) 5940ð¥3
Sol.9) Given expansion: (ð¥ â3
ð¥2)
ð
To find âmâ
General term: ðð+1 = (â1)ð mðð (ð¥)ðâð (3
ð¥2)2
= (â1)ð mðð (ð¥)ðâð 3ð
ð¥2ð
= ðð+1 = (â1)ð mðð (3)ð (ð¥)ðâ3ð For ð1 , put r = 0 = ð1 = (â1)0 mð0 (3)0 (ð¥)ð = ð1 = ð¥ð ⎠coefficient of ð1 = 1 For ð2 , put r = 1 = ð2 = (â1)1 mð1 3
1 ð¥ðâ3 = ð2 = -(ð)(3) ð¥ðâ3 ⎠coefficient of ð2 = -3m For ð3 put r =2 = ð3 = (â1)2 mð2 3
2 ð¥ðâ6 = ð3 = ðð2.9 ð¥ðâ6
= ð3 = 9ð(ðâ1)
2 ð¥ðâ6
⎠coefficient of ð3 = 9ð(ðâ1)
2
We are given that,
1 - 3m + 9ð(ðâ1)
2 = 559
â 2 - 6m + 9ð2 - 9m = 1118 â 9ð2 â 15m â 116 = 0 â 3m2 - 5m â 372 = 0 (divide by 3)
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m
a = 3, b = -5, c = -375 By quadratic formula,
m = 5屉25+(4)(3)(372)
2Ã3
m = 5屉4489
6
m = 5±67
6
m = 5+67
6 , m =
5â67
6
m = 72
6 , m =
â62
6
m= -12 (since power (m) cannot âve) ⎠general term becomes = ðð+1 = (â1)ð 12ðð (3)ð (ð¥)12â3ð For ð¥3 , put r =3 ⎠ð4 = (â1)3 12ð3 (3)3 ð¥3
= â12Ã11Ã10
6à 27 à ð¥3 = -5940ð¥3 ans.
Q.10) The coefficients of three consecutive terms in the expansion of (1 + ð)ð are in ratio 1:7:42. Find the value of ânâ?
mcqs a) 33 b) 26 c) 55 d) 78
Sol.10) Given expansion: (1 + ð)ð General term: ðð+1 = nðððð
Let the three consecutive terms are (ð â 1)ð¡â, (ð) ð¡âand (ð + 1)ð¡â term For ððâ1 , put r= r-2 ⎠ððâ1 = nððâ2ððâ2 Coefficient of ððâ1 = nððâ2 For ðð , put r= r-1 ⎠ðð = nððâ1ððâ1 Coefficient of ðð = nððâ1 ðð+1 = nðððð Coefficient of ðð+1 = nðð We are given that, nððâ2: nððâ1: nðð = 1:7:42
consider, nððâ2
nððâ1
= 1
7
â
ð!
(ðâ2)!(ðâð+2)!ð!
(ðâ1)!(ðâð+1)!
= 1
7
â (ðâ1)!(ðâð+1)!
(ðâ2)!(ðâð+2)! =
1
7
â (ðâ1)
(ðâð+2)! =
1
7
â 7r-7 = n-r+2 â 8r-9 = nâŠâŠâŠâŠ(1)
Now, consider nððâ1
nðð
= 7
42
â
ð!
(ðâ1)!(ðâð+1)!ð!
(ð)!(ðâð)!
= 1
6
â ð!(ðâð)!
(ðâ1)!(ðâð+1)! =
1
6
â ð(ðâ1)!(ðâð)!
(ðâ1)!(ðâð+1)(ðâð)! =
1
6
â ð
ðâð+1 =
1
6
â 6r + n â r +1 â 7r -1 = nâŠâŠâŠ(2)
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From (1) and (2), 8r-9 = 7r -1 â r=8, put in eq. (1) â n= 56 â 1 = 55 ans.
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