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Binomial Theorem

Class 11th Mathematics Bihar Board Solution
Exercise 8.1
  1. (1 - 2x)^5 Expand each of the expressions in Exercises 1 to 5.
  2. (2/x - x/2)^5 Expand each of the expressions in Exercises 1 to 5.…
  3. (2x - 3)^6 Expand each of the expressions in Exercises 1 to 5.
  4. (x/3 + 1/x)^5 Expand each of the expressions in Exercises 1 to 5.…
  5. (x + 1/x)^6 Expand each of the expressions in Exercises 1 to 5.
  6. (96)^3 Using binomial theorem, evaluate each of the following:
  7. (102)^5 Using binomial theorem, evaluate each of the following:
  8. (101)^4 Using binomial theorem, evaluate each of the following:
  9. (99)^5 Using binomial theorem, evaluate each of the following:
  10. Using Binomial Theorem, indicate which number is larger (1.1)^10000 or 1000.…
  11. Find (a + b)^4 - (a - b)^4 . Hence, evaluate (root 3 + root 2)^4 - (root 3 -…
  12. Find (x + 1)^6 + (x - 1)^6 . Hence or otherwise evaluate (root 2+1)^6 + (root…
  13. Show that 9n+1 - 8n - 9 is divisible by 64, whenever n is a positive integer.…
  14. Prove that sum _ r = 0^n3^rnc_r = 4^n
Exercise 8.2
  1. x^5 in (x + 3)^8 Find the coefficient of
  2. a^5 b^7 in (a - 2b)^12 . Find the coefficient of
  3. (x^2 - y)^6 Write the general term in the expansion of
  4. (x^2 - yx)^12 , x ≠ 0. Write the general term in the expansion of…
  5. Find the 4th term in the expansion of (x - 2y)^12 .
  6. Find the 13th term in the expansion of (9x - 1/3 root x)^18 , x not equal 0…
  7. (3 - x^3/6)^7 Find the middle terms in the expansions of
  8. (x/3 + 9y)^10 Find the middle terms in the expansions of
  9. In the expansion of (1 + a)m+n, prove that coefficients of am and an are equal.…
  10. The coefficients of the (r - 1)th, rth and (r + 1)th terms in the expansion of…
  11. Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the…
  12. Find a positive value of m for which the coefficient of x^2 in the expansion (1…
Miscellaneous Exercise
  1. Find a, b and n in the expansion of (a + b)n if the first three terms of the…
  2. Find a if the coefficients of x^2 and x^3 in the expansion of (3 + ax)^9 are…
  3. Find the coefficient of x^5 in the product (1 + 2x)^6 (1 - x)^7 using binomial…
  4. If a and b are distinct integers, prove that a - b is a factor of an - bn,…
  5. Evaluate (root 3 + root 2)^6 - (root 3 - root 2)^6
  6. Find the value of (a^2 + root a^2 - 1)^4 + (a^2 - root a^2 - 1)^4…
  7. Find an approximation of (0.99)^5 using the first three terms of its expansion.…
  8. Find n, if the ratio of the fifth term from the beginning to the fifth term from…
  9. Expand using Binomial Theorem (1 + x/2 - 2/x)^4 , x not equal 0
  10. Find the expansion of (3x^2 - 2ax + 3a^2)^3 using binomial theorem.…

Exercise 8.1
Question 1.

Expand each of the expressions in Exercises 1 to 5.

(1 – 2x)5


Answer:

We know that-



Hence






Thus,


Putting a = 1 & b = (-2x), we get,





Question 2.

Expand each of the expressions in Exercises 1 to 5.



Answer:

We know that-



Hence






Thus,


Putting a = (2/x) & b = (-x/2), we get-







Question 3.

Expand each of the expressions in Exercises 1 to 5.

(2x – 3)6


Answer:

We know that-



Hence






Thus,


Putting a = 2x & b = -3, we get-






Question 4.

Expand each of the expressions in Exercises 1 to 5.



Answer:

We know that-



Hence






Thus,


Putting a = (x/3) & b = (1/x), we get-








Question 5.

Expand each of the expressions in Exercises 1 to 5.



Answer:

We know that-



Hence






Thus,


Putting a = x & b = 1/x, we get-






Question 6.

Using binomial theorem, evaluate each of the following:

(96)3


Answer:

(96)3 = (100-4)3


We know that-



Hence






Thus,


Putting a = 100 & b = -4, we get-






Thus,



Question 7.

Using binomial theorem, evaluate each of the following:

(102)5


Answer:

(102)5 = (100+2)5


We know that-



Hence






Thus,


Putting a = 100 & b = 2, we get-






Hence, (102)5 = 11040808032



Question 8.

Using binomial theorem, evaluate each of the following:

(101)4


Answer:

(101)4 = (100+1)4


We know that-



Hence







Thus,


Putting a = 100 & b = 1, we get-






Thus,



Question 9.

Using binomial theorem, evaluate each of the following:

(99)5


Answer:

(99)5 = (100-1)5


We know that-



Hence






Thus,


Putting a = 100 & b = -1, we get-









Hence, (99)5 = 9509900499



Question 10.

Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.


Answer:

(1.1)10000 = (1+0.1)10000


We know that-



Hence






Putting a = 1 & b = 0.1, we get-






Hence, (1.1)10000 is larger than 1000.



Question 11.

Find (a + b)4 – (a – b)4. Hence, evaluate


Answer:

We know that-



Hence







Thus,


Replacing b with -b




Now,






Putting in , we get-






Question 12.

Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate .


Answer:

We know that-



Hence






Thus,


Replacing b with -b





Now,





Putting a = x & b = 1 in , we get-




putting x = √2






Hence,



Question 13.

Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer.


Answer:

In order to show that 9n+1 – 8n – 9 is divisible by 64,


we have to prove that


9 n+1 – 8n – 9 = 64 k, where k is some natural number

Now,

9n+1 = (1+8)n+1


We know that-



putting a =1, b = 8, and n = n+1










Hence,



Taking out (8)2 from right side, we get-





where is a natural number


Thus, 9n+1 – 8n – 9 is divisible by 64.


Hence Proved.


Question 14.

Prove that


Answer:

By Binomial Theorem

On right side we need 4n so we will put the values as,

Putting b = 3 & a = 1 in the above equation, we get-





Hence Proved.



Exercise 8.2
Question 1.

Find the coefficient of

x5 in (x + 3)8


Answer:

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br


Here x5 is the Tr+1 term so


a= x, b = 3 and n =8


Tr+1 = 8Cr x8-r 3r……………1


For finding out x5


We equate x5= x8-r


⇒ r= 3


Putting value of r in 1 we get


T3+1 = 8C3 x8-3 33



= 1512 x5


Hence the coefficient of x5= 1512



Question 2.

Find the coefficient of

a5b7 in (a – 2b)12 .


Answer:

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here a =a , b = -2b & n =12


Putting values


Tr+1 = 12Cr a12-r (-2b)r……….1


To find a5


We equate a12-r =a5


r=7


Putting r = 7 in 1


T8 = 12C7 a5 (-2b)7



= -101376 a5 b7


Hence the coefficient of a5b7= -101376



Question 3.

Write the general term in the expansion of

(x2 – y)6


Answer:

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br . ……..1

Here a = x2 , n =6 and b = -y


Putting values in 1


Tr+1 = 6Cr x 2(6-r) (-1)r yr





Question 4.

Write the general term in the expansion of

(x2 – yx)12, x ≠ 0.


Answer:

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here n = 12, a= x2 and b = -yx


Putting the values


Tn+1 =12Cr × x2(12-r) (-1)r yr xr


=


=



Question 5.

Find the 4th term in the expansion of (x – 2y)12.


Answer:

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here a= x , n =12 , r= 3 and b = -2y


T4 = 12C3 x9 (-2y)3


=


=


= -1760 x9 y3



Question 6.

Find the 13th term in the expansion of


Answer:

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here a=9x , , n =18 and r = 12


Putting values



= (9 = (32)6 = 312)


= 18564



Question 7.

Find the middle terms in the expansions of



Answer:

Here n = 7 so there would be two middle terms given by


Now


For T4 , r= 3


The term will be


Tr+1 = nCr an-r br





For T5 term , r = 4


The term Tr+1 in the binomial expansion is given by


Tr+1 = nCr an-r br





Question 8.

Find the middle terms in the expansions of



Answer:

Here n is even so the middle term will be given by

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br



Putting the values



=


= 61236 x5y5



Question 9.

In the expansion of (1 + a)m+n, prove that coefficients of am and an are equal.


Answer:

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here n= m+n , a = 1 and b= a


Putting the values in the general form


Tr+1 = m+nCr 1m+n-r ar


= m+nCr ar………….1

Now we have that the general term for the expression is,

Tr+1 = m+nCr ar
Now, For coefficient of am

Tm+1 = m+nCm am
Hence, for coefficient of am, value of r = m

So, the coefficient is m+nCm

Similarly, Coefficient of an is m+nCn

m+nCm =

And also,

m+nCn =

The coefficient of am and an are same i.e.;


Hence proved.


Question 10.

The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1 : 3 : 5. Find n and r.


Answer:

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here the binomial is (1+x)n with a = 1 , b = x and n = n


The (r+1)th term is given by


T(r+1) = nCr 1n-r xr


T(r+1) = nCr xr


The coefficient of (r+1)th term is nCr


The rth term is given by (r-1)th term


T(r+1-1) = nCr-1 xr-1


Tr = nCr-1 xr-1


∴ the coefficient of rth term is nCr-1


For (r-1)th term we will take (r-2)th term


Tr-2+1 = nCr-2 xr-2


Tr-1 = nCr-2 xr-2


∴ the coefficient of (r-1)th term is nCr-2


Given that the coefficient of (r-1)th, rth and r+1th term are in ratio


1:3:5









⇒ 3r - 3 = n – r + 2


⇒ n - 4r + 5 =0…………1


Also









⇒ 5r = 3n - 3r + 3


⇒ 8r – 3n - 3 =0………….2


We have 1 and 2 as


n - 4r 5 =0…………1


8r – 3n - 3 =0…………….2


Multiplying equation 1 by number 2


2n -8r +10 =0……………….3


Adding equation 2 and 3


2n -8r +10 =0


+ -3n – 8r - 3 =0


⇒ -n = -7


n =7 and r = 3


Question 11.

Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n – 1.


Answer:

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

The general term for binomial (1+x)2n is


Tr+1 = 2nCr xr …………………..1


To find the coefficient of xn


r=n


Tn+1 = 2nCn xn


The coefficient of xn = 2nCn


The general term for binomial (1+x)2n-1 is


Tr+1 = 2n-1Cr xr


To find the coefficient of xn


Putting n =r


Tr+1 = 2n-1Cr xn


The coefficient of xn = 2n-1Cn


We have to prove


Coefficient of xn in (1+x)2n = 2 coefficient of xn in (1+x)2n-1



Hence L.H.S = R.H.S.



Question 12.

Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.


Answer:

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here a = 1, b = x and n = m


Putting the value


Tr+1 = mCr 1m-r xr


= mCr xr


We need coefficient of x2


∴ putting r = 2


T2+1 = mC2 x2


The coefficient of x2 = mC2


Given that coefficient of x2 = mC2 = 6




⇒ m(m-1) = 12


⇒ m2- m - 12 =0


⇒ m2- 4m +3m - 12 =0


⇒ m(m-4) + 3(m-4) = 0


⇒ (m+3)(m - 4)= 0


⇒ m = - 3, 4


we need positive value of m so m = 4




Miscellaneous Exercise
Question 1.

Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively.


Answer:

We know that-



So first 3 terms are


Also, it is given that their value are 729, 7290 and 30375


Therefore,


…(1)


…(2)


…(3)


Dividing (1) by (2)







…(4)


Dividing (2) by (3)








…(5)


Dividing (4) by (5)





⇒ 12(n-1) = 10n


⇒ 12n-12 = 10n


⇒ 2n = 12


⇒ n = 12/2


∴ n = 6


Putting n = 6 in (1)


an = 729


⇒ a6 = 729


⇒ a6 = (3)6


∴ a = 3


Putting a = 3, n = 6 in (5)




⇒ 6b = 30


⇒ b = 30/6


∴ b = 5


Thus, a = 3, b = 5 and n = 6.



Question 2.

Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.


Answer:

We know that-


General term of expansion (a+b)n is



For (3+ax)9


Putting a = 3, b = ax & n = 9


General term of (3+ax)9 is




Since we need to find the coefficients of x2 and x3, therefore


For r = 2



Thus, the coefficient of x2 =


For r = 3



Thus, the coefficient of x3 =


Given that-


Coefficient of x2 = Coefficient of x3







∴ a = 9/7


Hence, a = 9/7



Question 3.

Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.


Answer:

We know that-



Hence





= a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6


Thus, (a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6


Putting a = 1 & b = 2x, we get-


(1+2x)6 = (1)6 + 6(1)5(2x) + 15(1)4(2x)2 + 20(1)3(2x)3


+ 15(1)2(2x)4 + 6(1)(2x)5 + (2x)6


= 1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6


Similarly,







= a7 + 7a6b + 21a5b2 + 35a4b3 + 35a3b4 + 21a2b5 + 7ab6 +b7


Thus, (a + b)7 = a7 + 7a6b + 21a5b2 + 35a4b3 + 35a3b4 + 21a2b5 + 7ab6 +b7


Putting a = 1 & b = -x, we get-


(1-x)7 = (1)7 + 7(1)6(-x) + 21(1)5(-x)2 + 35(1)4(-x)3


+ 35(1)3(-x)4 + 21(1)2(-x)5 + 7(1)(-x)6 + (-x)7


= 1 - 7x + 21x2 - 35x3 + 35x4 - 21x5 + 7x6 - x7


Now,


(1+2x)6(1-x)7


=(1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6)


× (1 - 7x + 21x2 - 35x3 + 35x4 - 21x5 + 7x6 - x7)


Coefficient of x5 = [(1)×(-21) + (12)×(35) + (60)×(-35)


+ (160)×(21) + (240)×(-7) + (192)×(1)]


= [-21 + 420 - 2100 + 3360 - 1680 + 192]


= 171


Thus, the coefficient of x5 in the expression (1+2x)6(1-x)7 is 171.



Question 4.

If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer. [Hint write an = (a – b + b)n and expand]


Answer:

We can write an as


an = (a-b+b)n


We know that-



putting a = b & b = a-b, we get-












where


Hence (a-b) is a factor of (an-bn).



Question 5.

Evaluate


Answer:

We know that-



Hence





= a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6


Thus, (a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6


Replacing b with -b


(a +(-b))6 = a6 + 6a5(-b) + 15a4(-b)2 + 20a3(-b)3 + 15a2(-b)4 + 6a(-b)5 + (-b)6


(a - b)6 = a6 - 6a5b + 15a4b2 - 20a3b3 + 15a2b4 - 6ab5 + b6


Now,


(a + b)6 - (a - b)6


= (a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6)


- (a6 - 6a5b + 15a4b2 - 20a3b3 + 15a2b4 - 6ab5 + b6)


= 2(6a5b + 20a3b3 + 6ab5)


Putting a = √3 & b = √2, we get-


(√3 + √2)6 - (√3 - √2)6


= 2[6(√3)5(√2) + 20(√3)3(√2)3 + 6(√3)(√2)5]


= 2[6(9√3)(√2) + 20(3√3)(2√2) + 6(√3)(4√2)]


= 2[54√6 + 120√6 + 24√6]


= 2[198√6]


= 396√6



Question 6.

Find the value of


Answer:

We know that-



Hence






= a4 + 4a3b + 6a2b2 + 4ab3 + b4


Thus, (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4


Replacing b with -b


(a + (-b))4 = a4 + 4a3(-b) + 6a2(-b)2 + 4a(-b)3 + (-b)4


(a-b)4 = a4 - 4a3b + 6a2b2 - 4ab3 + b4


Now,


(a + b)4 + (a - b)4


= (a4 + 4a3b + 6a2b2 + 4ab3 + b4) + (a4 - 4a3b + 6a2b2 - 4ab3 + b4)


= 2(a4 + 6a2b2 + b4)


Putting a = a2 & b = √(a2-1)


(a2 + √(a2-1))4 + (a2 - √(a2-1))4


= 2[(a2)4 + 6(a2)2(√(a2-1))2 + (√(a2-1))4]


= 2[ a8 + 6(a4)(a2-1) + (a2-1)2]


= 2[ a8 + 6 a6 - 6a4 + a4 - 2 a2 + 1]


= 2[ a8 + 6 a6 - 5a4 - 2 a2 + 1]



Question 7.

Find an approximation of (0.99)5 using the first three terms of its expansion.


Answer:

(0.99)5 = (1 - 0.01)5


We know that-



Hence





= a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5


Thus, (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5


Putting a = 1 & b = -0.01, we get-


(1 + (-0.01))5 = (1)5 + 5(1)4(-0.01) + 10(1)3(-0.01)2 + 10(1)2(-0.01)3 + 5(1)(-0.01)4 + (-0.01)5


Using first three terms,


(0.99)5 = (1)5 + 5(1)4(-0.01) + 10(1)3(-0.01)2


= 1 - 0.05 + 0.001


= 1.001 - 0.050


= 0.951



Question 8.

Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of is √6 : 1


Answer:

We know that


General term of expansion (a + b)n



We need to calculate fifth term from beginning of expansion



∴ putting r = 4, a = , and b = , we get-





Now,


In the expression of (a + b)n


rth term from the end = (n-r+2)th term from the begining


Hence, 5th term from the end


= (n-5+2)th term from the beginning


= (n-3)th term from the beginning


Now, We need to calculate (n-3)th term from beginning of expansion



putting r = (n-3)-1 = n-4, a = , and b = , we get-






Given that-









Comparing powers of 6



⇒ 2(n-8) = 4


⇒ 2n-16 = 4


⇒ 2n = 20


∴ n = 20/2 = 10


Thus, the value of n is 10.



Question 9.

Expand using Binomial Theorem


Answer:

We know that-



Hence






= a4 + 4a3b + 6a2b2 + 4ab3 + b4


Thus, (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 …(1)


Putting , we get-



…(2)


Now Solving separately


From (1)


(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4


putting a = 1 & b = (x/2), we get-





we know that-


(a + b)3 = a3 + 3a2b + 3ab2 + b3


putting a = 1 & b = (x/2), we get-




Substituting the value of in (2), we get-







Thus,



Question 10.

Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem.


Answer:

We know that-


(a + b)3 = a3 + 3a2b + 3ab2 + b3


putting a = 3x2 & b = -a(2x-3a), we get-


[3x2 + (-a(2x-3a))]3


= (3x2)3+3(3x2)2(-a(2x-3a)) + 3(3x2)(-a(2x-3a))2 + (-a(2x-3a))3


= 27x6 - 27ax4(2x-3a) + 9a2x2(2x-3a)2 - a3(2x-3a)3


= 27x6 - 54ax5 + 81a2x4 + 9a2x2(4x2-12ax+9a2)


- a3[(2x)3 - (3a)3 - 3(2x)2(3a) + 3(2x)(3a)2]


= 27x6 - 54ax5 + 81a2x4 + 36a2x4 - 108a3x3 + 81a4x2


-8a3x3 + 27a6 + 36a4x2 - 54a5x


= 27x6 - 54ax5+ 117a2x4 - 116a3x3 + 117a4x2 - 54a5x + 27a6


Thus, (3x2 – 2ax + 3a2)3


= 27x6 - 54ax5+ 117a2x4 - 116a3x3 + 117a4x2 - 54a5x + 27a6