NCERT Solutions Class 11 Mathematics Binomial Theorem Exercise 8.2

Question 9:
In the expansion of (1 + a)^{m + n}, prove that coefficients of a^m and aⁿ are equal.

Answer:
It is known that (r + 1)^th term, (T_r+1), in the binomial expansion of (a + b)ⁿ is given by
T_r+1 = ⁿC_r a^n-r b^r
Assuming that a^m occurs in the (r + 1)^th term of the expansion (1 + a)^{m + n}, we obtain
T_r+1 = ^m+nC_r (1)^m+n-r a^r
Comparing the indices of a in a^m and in T_{r + 1}, we obtain
r = m
Therefore, the coefficient of a^m is
^m+nC_m = (m + n)!/{m! * (m +n – m)!} = (m + n)!/(m! * n!)   ……………1
Assuming that aⁿ occurs in the (k + 1)^th term of the expansion (1 + a)^m+n, we obtain
T_k+1 = ^m+nC_k (1)^m+n-k a^k = ^m+nC_k a^k
Comparing the indices of a in aⁿ and in T_{k + 1}, we obtain
k = n
Therefore, the coefficient of aⁿ is
^m+nC_n = (m + n)!/{n! * (m +n – n)!} = (m + n)!/(n! * m!)   ……………2
Thus, from equation 1 and 2, it can be observed that the coefficients of a^m and aⁿ in the
expansion of (1 + a)^{m + n} are equal.

Question 10:
The coefficients of the (r – 1)^th, r^th and (r + 1)^th terms in the expansion of (x + 1)ⁿ are in the ratio 1 : 3 : 5. Find n and r.

Answer:
It is known that (k + 1)^th term, (T_k+1), in the binomial expansion of (a + b)ⁿ is given by
T_k+1 = ⁿC_k a^n-k b^k
Therefore, (r – 1)^th term in the expansion of (x + 1)ⁿ is
T_r-1 = ⁿC_r-2 (x)^{n-(r – 2)} 1^r-2 = ⁿC_r-2 (x)^{n – r + 2}
(r + 1) term in the expansion of (x + 1)ⁿ is
T_r+1 = ⁿC_r (x)^n-r  1^r = ⁿC_r (x)^{n – r}  
r^th term in the expansion of (x + 1)ⁿ is
T_r = ⁿC_r-1 (x)^{n-(r – 1)} 1^r-1 = ⁿC_r-1 (x)^{n – r + 1}
Therefore, the coefficients of the (r – 1)^th, r^th, and (r + 1)^th terms in the expansion of (x + 1)ⁿ are
respectively. Since these coefficients are in the ratio 1:3:5, we obtain
ⁿC_r-2 / ⁿC_r-1 = 1/3 and ⁿC_r-1 / ⁿC_r = 3/5
Now, ⁿC_r-2 / ⁿC_r-1 = 1/3
=> [n!/{(r – 2)! * (n – r + 2)!}]/[n!/{(r – 1)! * (n – r + 1)!}] = 1/3
=> [(r – 1) * (r – 2)! * (n – r + 1)!]/ [(r – 2) * (n – r + 2)! * (n – r + 1)!] = 1/3
=> (r – 1)/(n – r + 2) = 1/3
=> 3(r – 1) = (n – r + 2)
=> 3r – 3 = n – r + 2
=> n – 4r + 5 = 0  …………1
Again, ⁿC_r-1 / ⁿC_r = 3/5
=> [n!/{(r – 1)! * (n – r + 1)!}]/[n!/{r! * (n – r)!}] = 3/5
=> [r * (r – 1)! * (n – r)!]/ [(r – 1) * (n – r + 1)! * (n – r)!] = 3/5
=> r/(n – r + 1) = 3/5
=> 5r = 3(n – r + 1)
=> 5r = 3n – 3r + 3
=> 3n – 8r + 3 = 0  …………2
Solving equations 1 and 2, we get
Multiplying equation 1 by 3 and subtracting it from equation 2, we obtain
     4r – 12 = 0
=> r = 3
Putting the value of r in equation 1, we obtain
      n – 12 + 5 = 0
=> n = 7
Thus, the values of n and r are 7 and 3 respectively.

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:
It is known that (r + 1)^th term, (T_r+1), in the binomial expansion of (a + b)ⁿ is given by
T_r+1 = ⁿC_r a^n-r b^r
Assuming that xⁿ occurs in the (r + 1)^th term of the expansion of (1 + x)²ⁿ, we obtain
T_r+1 = ²ⁿC_r (1)^{2n – r}  x^r = ²ⁿC_r x^r  
Comparing the indices of x in xⁿ and in T_{r + 1}, we obtain r = n
Therefore, the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is
²ⁿC_n = (2n)!/{n! * (2n – n)!} = (2n)!/{n! * n!} = (2n)!/(n!)²   ……….1
Assuming that xⁿ occurs in the (k +1)^th term of the expansion (1 + x)^{2n – 1}, we obtain
T_k+1 = ^2n-1C_k (1)^2n-1-k  x^k = ^2n-1C_k (x)^k  
Comparing the indices of x in xⁿ and T_{k + 1}, we obtain k = n
Therefore, the coefficient of xⁿ in the expansion of (1 + x)^{2n –1} is
^2n-1C_n = (2n – 1)!/{n! * (2n – 1 – n)!}
          = (2n – 1)!/{n! * (n – 1)!}
      = {2n * (2n – 1)!/{2n * n! * (n – 1)!}
      = (2n)!/(2 * n! * n!)  
      = (2n)!/{2 * (n!)²} ……….2
From equation 1 and 2, it is observed that
         ²ⁿC_n /2 = ^2n-1C_n
=> ²ⁿC_n = 2(^2n-1C_n)
Therefore, 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.
Hence, proved.

Question 12:
Find a positive value of m for which the coefficient of x² in the expansion (1 + x)^m is 6.

Answer:
It is known that (r + 1)^th term, (T_r+1), in the binomial expansion of (a + b)ⁿ is given by
T_r+1 = ⁿC_r a^n-r b^r
Assuming that x² occurs in the (r + 1)^th term of the expansion (1 +x)^m, we obtain
T_r+1 = ^mC_r (1)^{m – r}  x^r = ^mC_r (x)^r  
Comparing the indices of x in x² and in T_{r + 1}, we obtain r = 2
Therefore, the coefficient of x² is ^mC₂
It is given that the coefficient of x² in the expansion (1 + x)^m is 6.
So, ^mC₂ = 6
=> m!/{2! * (m – 2)!} = 6
=> {m(m – 1)(m – 2)!}/{2 * (m – 2)!} = 6
=> {m(m – 1)}/2 = 6
=> m(m – 1) = 12
=> m² – m – 12 = 0
=> (m – 4)(m + 3) = 0
=> m = -3, 4
Thus, the positive value of m, for which the coefficient of x² in the expansion (1 + x)^m is 6, is 4