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BINOMIAL THEOREM Flipbook PDF
PART-II
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BINOMIAL THEOREM MANU
December 19, 2020
MANU
BINOMIAL THEOREM
General Term
MANU
BINOMIAL THEOREM
QUESTION01 Consider the expansion of (3x 2 −
1 9 ) 3x
1
Find the coeffiecent of x 6
2
Find the term independant of x
ANSWER01 (r + 1)th term = (−1)r nCr an−r b r a = 3x 2 , b =
1 ,n 3x
=9
1 r (r + 1)th term = (−1)r 9Cr (3x 2 )9−r ( 3x ) 9−r
(r + 1)th term = (−1)r 9Cr 3 3r
x 18−2r xr
= (−1)r 9Cr 39−r −r x 18−2r −r = (−1)r 9Cr 39−2r x 18−3r To find the coeffiecent of x 6 exponent of x must be 6 So take 18-3r=6,∴, 3r = 18 − 6 = 12, r = Coeffiecent of
x6
=
(−1)49C4 39−8
=
39C4
12 3
=4
= 3 × 126 = 378
To find term independant of x exponent of x must be zero take 18-3r=0,So r =
18 3
=6
Term independant of x = (−1)69C6 39−12 =
9.8.7 1 1.2.3 27
=
28 9 MANU
BINOMIAL THEOREM
QUESTION02 1
Find the number of terms in the expansion of (x − x1 )14
2
Find the general term in the expansion of (x − x1 )14
3
Find the term independent of x in the above expansion
Answer02 (1) 15 (2) (r + 1)th term = (−1)r nCr an−r b r a = x, b =
1 ,n x
= 14
(r +
1)th
term = (−1)r 14Cr x 14−r ( x1 )r
(r +
1)th
term = (−1)r 14Cr x 14−r −r
(r +
1)th
term = (−1)r 14Cr x 14−2r
(3) To find term independant of x exponent of x must be zero take 14-2r=0,So r =
14 2
=7
Term independant of x = (−1)714C7 = −14C7
MANU
BINOMIAL THEOREM
QUESTION03 Consider expansion of (x 3 + x1 )8 1
Find the general term
2
Find the term independent of x in the above expansion
3
Find the coeffiecent of x 8
Answer03 (r + 1)th term = nCr an−r b r a = x 3, b = (r +
1)th
1 ,n x
=8
term = 8Cr (x 3 )8−r ( x1 )r
(r + 1)th term = 8Cr x
24−3r xr
(r + 1)th term = 8Cr x 24−3r −r (r + 1)th term = 8Cr x 24−4r To find term independant of x exponent of x must be zero take 24-4r=0,So r =
24 4
=6
Term independant of x = 8C6 = 8C2 = 28 To find the coeffiecent of x 8 exponent of x must be 8 take 24-4r=8,4r=24-8,So r = Coeffiecent of
x8
=
16 4
=4
8C 4 MANU
BINOMIAL THEOREM
QUESTION04 Consider the expansion of (x 2 −
1 9 ) 3x
1
Find the coeffiecent of x 9
2
Find the term independant of x
ANSWER04 (r + 1)th term = (−1)r nCr an−r b r a = x 2, b = (r +
1)th
1 ,n 3x
=9
1 r term = (−1)r 9Cr (x 2 )9−r ( 3x )
(r + 1)th term = (−1)r 9Cr 31r
x 18−2r xr
= (−1)r 9Cr 31r x 18−2r −r = (−1)r 9Cr 31r x 18−3r To find the coeffiecent of x 6 exponent of x must be 9 So take 18-3r=9,∴, 3r = 18 − 9 = 9, r =
9 3
=3
Coeffiecent of x 6 = (−1)3 9C3 313 = − 84 27 To find term independant of x exponent of x must be zero take 18-3r=0,So r =
18 3
=6
Term independant of x = (−1)6 9C6 316 =
9.8.7 1 1.2.3 729
=
28 43 MANU
BINOMIAL THEOREM
QUESTION05 Consider the expansion of (x 2 − x2 )18 1
Determine whether the expansion contain a term of x 10
2
Find the term independant of x
ANSWER05 (r + 1)th term = (−1)r nCr an−r b r a = x 2, b =
2 ,n x
= 18
(r + 1)th term = (−1)r
18C (x 2 )18−r ( 2 )r r x
(r + 1)th term = (−1)r 2r
18C x 36−2r r xr
18C x 36−2r −r r r 2 18Cr x 36−3r
= (−1)r 2r =
(−1)r
To find the coeffiecent of x 10 exponent of x must be 10 ∴, 36 − 3r = 10, r = term of x 10
26 3
is not an integer So the given expansion not conatining a
To find term independent of x exponent of x must be zero take 36-3r=0,So r =
36 3
= 12
Term independent of x = (−1)12 212 =
18C 12
212 18C6 MANU
BINOMIAL THEOREM
QUESTION06 1
The number of terms in the expansion of ( x3 + 9y )10 is.....................
2
Find the middle term in the above expansion
ANSWER06 (1) 11 (2) Since n=10 is even Middle term =( 10 + 1)th term = 6th term 2 (r + 1)th term =
nC an−r b r r
r + 1 = 6, r = 6 − 1 = 5 a= 6th
x ,b 3
= 9y , n = 10
term =
10C ( x )10−5 (9y )5 5 3
Middle term =
95 10 C5 x 5 y 5 35
= 35 =
10C x 5 y 5 5 243 10C5 x 5 y 5
MANU
BINOMIAL THEOREM
QUESTION07 2
1
The number of terms in the expansion of ( 3x2 −
2
Find the middle term in the above expansion
3
Find term independent of x
1 6 ) 3x
is.....................
ANSWER07 (1) 7 (2) Since n=6 is even Middle term =( 26 + 1)th term = 4th term (r + 1)th term = −1)r nCr an−r b r r + 1 = 4, r = 4 − 1 = 3 a= 4th
3x 2 ,b 2
term =
=
1 ,n 3x
=6
(−1)3 6C
3x 2 6−3 1 3 ( 3x ) 3( 2 )
6 33 6 C3 xx 3 23 33 − 25 x 3
Middle term = − =−
20 8
x3 =
2
1 r (r + 1)th term = (−1)r 6Cr ( 3x2 )6−r ( 3x )
(r + 1)th term = (−1)r
12−2r 36−r 6 Cr x x r 26−r 3r
= (−1)r
36−r −r 6 Cr x 12−2r −r 26−r
= (−1)r
36−2r 6 Cr x 12−3r 26−r MANU
BINOMIAL THEOREM
ANSWER07...... To find term independent of x exponent of x must be zero take 12-3r=0,So r =
12 3
=4
Term independent of x = =
15 32 22
=
15 36
=
36−8 6 C4 26−4
5 12
MANU
BINOMIAL THEOREM
QUESTION08 Consider the expansion of (x + x1 )10 2
1
The number of terms in the expansion of ( 3x2 −
2
Find the middle term in the above expansion
1 6 ) 3x
is.....................
ANSWER08 (1) 11 (2) Since n=10 is even Middle term =( 10 + 1)th term = 6th term 2 (r + 1)th term =
nC an−r b r r
r + 1 = 6, r = 6 − 1 = 5 a = x, b = 6th term =
1 , n = 10 x 10C (x)10−5 ( 1 )5 5 x
Middle term =
10C
x 5 5( x )
= 10C5
MANU
BINOMIAL THEOREM
QUESTION09 Consider the expansion of ( x2 − x2 )10 1
Find the general term
2
Find the middle term
ANSWER09 1) (r + 1)th term = (−1)r nCr an−r b r x , b = x2 , n = 10 2 + 1)th term = (−1)r 10Cr ( x2 )10−r ( x2 )r
a= (r
2r 10C x 10−r r xr 210−r x 10−r −r
(r + 1)th term = (−1)r = (−1)r 2r −10+r = (−1)r 22r −10
10C r 10C x 10−2r r
(2) Since n=10 is even Middle term =( 10 + 1)th term = 6th term 2 r+1=6,r=6-1=5 = (−1)5 210−10 =
−10C
10C x 10−10 5
5
MANU
BINOMIAL THEOREM