AP Inter 1st Year Maths Exercise 7c Solutions

Referring to the AP Inter 1st Year Maths Study Material Chapter 7 Binomial Theorem Exercise 7c Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Binomial Theorem Solutions Exercise 7c

Find the coefficient of

Question 1.
x⁵ in (x + 3)⁸
Solution:
The general term T_r+1 of the binomial expansion (a + b)n is given by T_r+1 = ⁿC_r a^n-rb^r
Here x5 is the T_r+1 term so a = x, b = 3 and n = 8
T_r+1 = 8C_rx^8-r3^r ………..(1)
To find out the coefficient of x⁵, We have to equate x⁵ = x^8-r =>r = 3
Putting the value of r in (1), we get T₃₊₁ = ⁸C₃ × x^8-3 3³
T₄ = \(\frac{8!}{3!5!}\) × x⁵ × 27 = 1512 x⁵
Hence the coefficient of x⁵ = 1512

Question 2.
a⁵b⁷ in (a – 2b)¹².
Solution:
The general term Tr+ t of the binomial expansion (a + b)ⁿ is given by T_r+1 = ⁿC_r a^n-rb^r
Here a = a, b = -2b & n = 12
Substituting above values, we get T_r+1 = ¹²C_r a^12-r (-2b)^r … (1)
To find out the coefficient of a⁵, We equate a^12-r = a⁵, r = 7
Putting r = 7 in (1) ⇒ Tg = ¹²C₇ a⁵(-2b)⁷
T₈ = \(\frac{12!}{7!5!}\) × a⁵ × (-2)⁷b⁷
= -101376 a⁵b⁷
Hence, the coefficient of a⁵b⁷ = -101376.

Write the general term in the expansion of

Question 3.
(x³ – y)⁶
Solution:
The general term T_r+1 of the binomial expansion (a + b)n is given by T_r+1 = nCra”-b’… (1)
Here, a = x², n = 6 and b = – y
Putting above values in (1)
T_r+1 = ⁶C_r x^2(6-r) (-1)^r y^r
= ⁶C_r x^12-2r x (-1)^r X y^r
= (-1)^{r 6}C_r x^12-2r y^r

Question 4.
(x² – yx)¹², x ≠ 0.
Solution:
The general term T_r+x of the binomial expansion (a + b)n is given by T_r+1 = nC_r a^n-r b^r
Here n = 12, a = x² and b = -y x
Substituting the above values, we get
T_r+1 = ¹²C_r x^2(12-r) (-1)^r y^rx^r = (-1)r ¹²C_r x^24-2r y^r x^r
= (-1)^{r 12}C_r x^24-r. y^r

Question 5.
In the expansion of (1 + a)^m+n, prove that the coefficients of am and an are equal.
Solution:
We know that the general term T_r+1 of the binomial expansion (a + b)ⁿ is given by
T_r+1 = ⁿC_r a^n-r b^r
Here n = m + n, a = 1 and b = a
Substituting the values in the general term
T_r+1 = ^m+nC. 1^m+n-ra^r = ^m+nC_r a^r ………..(1)
Now, we have the general term for the expression is, T_r+1 = ^m+nC_r a^r
Now, for coefficient of a^m, T_m+1 = ^m+nC_m a^m
Hence, for the coefficient of m, the value of r = m
So, the coefficient is ^m+nC_m
Similarly, the coefficient of an is ^m+nC_n
^m+nC_m = \(\frac{(\mathrm{m}+\mathrm{n})!}{\mathrm{m}!\mathrm{n}!}\) And also, ^m+nC_n = \(\frac{(\mathrm{m}+\mathrm{n})!}{\mathrm{m}!\mathrm{n}!}\)
The coefficient of am and an are same that is \(\frac{(\mathrm{m}+\mathrm{n})!}{\mathrm{m}!\mathrm{n}!}\)

Question 6.
The coefficients of the (r – 2)^th, r^th and (r + 2)^th terms in the expansion of (x + 1)ⁿ are in the ratio 1:3:5. Find n and r.
Solution:
The general term T_r+1 of the binomial expansion (a + b)n is given by T_r+1 = ⁿC_r a^n-r b^r
Here, the binomial is (1 + x)ⁿ where a = 1, b = x and n = n

The (r + 1)^th term is given by
T_r+1 = ⁿC_r 1^n-r x_r ⇒ T_{(r + 1)} = ⁿC_r x^r
The coefficient of (r + 1)^th term is ⁿC_r

The rth term is given by (r – 1)^th term
T_{(r – 1 – 1)} = ⁿC_r-1
⇒ T_r = ⁿC_{r – 1}x^r-1
∴ The coefficeint of (r – 1)^th term is ⁿC_{r – 1}

For (r – 1)^th term, we will take (r – 2)^th term
T_{r – 2 + 1} = ⁿC_r-2 x^r-2
T_r-1 = ⁿC_r-2x^r-2
∴ The coefficeint of (r – 1)^th term is ⁿC_{r – 2}

Given that the coefficient of (r – 1)^th, r^th and r + 1^th term are in ratio 1: 3 : 5
Therefore,
\(\frac{\text { Coefficient of }(\mathrm{r}-1)^{\text {th }} \text { term }}{\text { Coefficient of } \mathrm{r}^{\text {th }} \text { term }}=\frac{1}{3}\)

\(\frac{{ }^n C_{r-2}}{{ }^n C_{r-1}}=\frac{1}{3} \Rightarrow \frac{\frac{n!}{(r-2)!(n-r+2)!}}{\frac{n!}{(r-1)(n-r+1)!}}=\frac{1}{3}\)

On rearrangement, we get
\(\frac{n!}{(r-2)!(n-r+2)!} \times \frac{(r-1)!(n-r+1)!}{n!}=\frac{1}{3}\)

By multiplying
⇒ \(\frac{(\mathrm{r}-1)(\mathrm{r}-2)!(\mathrm{n}-\mathrm{r}+1)!}{(\mathrm{r}-2)!(\mathrm{n}-\mathrm{r}+2)!}=\frac{1}{3} \Rightarrow \frac{(\mathrm{r}-1)(\mathrm{n}-\mathrm{r}+1)!}{(\mathrm{n}-\mathrm{r}+2)(\mathrm{n}-\mathrm{r}+1)!}=\frac{1}{3}\)

On simplifying we get,
⇒ \(\frac{(\mathrm{r}-1)}{(\mathrm{n}-\mathrm{r}+2)}=\frac{1}{3}\)
⇒ 3r – 3 = n – r + 2
⇒ n – 4r + 5 = 0 …. (1)

Also, \(\frac{\text { the coefficient of } r^{\text {th }} \text { term }}{\text { coefficient of } r+1^{\text {th }} \text { term }}=\frac{3}{5} \Rightarrow \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{3}{5}\)

On rearranging we get,
⇒ \(\frac{n!}{(r-1)!(n-r+1)!} \times \frac{r!(n-r)!}{n!}=\frac{3}{5}\)

By multiplying
⇒ \(\frac{r(r-1)!(n-r)!}{r-1)!(n-r+1)!}=\frac{3}{5} \Rightarrow \frac{r(n-r)!}{(n-r+1)!}=\frac{3}{5} \Rightarrow \frac{r(n-r)!}{(n-r+1)(n-r)!}=\frac{3}{5}\)

On simplifying we get,
⇒ \(\frac{r}{(n-r+1)}=\frac{3}{5}\)
⇒ 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 equations (2) and (3)
2n – 8r + 10 = 0
⇒ -3n – 8r – 3 = 0
⇒ -n = -7
⇒ n = 7 and r = 3

Question 7.
Prove that the coefficient of xn in the expansion of (1 + x)²ⁿ is twice the of xⁿ in the expansion of (1 + x)^2n-1.
Solution:
The general term T_r+1, of the binomial expansion (a + b)ⁿ is given by T_r+1 = ⁿC_r+1 a^n-rb^r
The general term for binomial (1 + x)²ⁿ is
T_(r+1) = ²ⁿC_rx^r …(1)
To find the coefficient of xⁿ, r = n
T_n+1 = ²ⁿC_nxⁿ
The coefficeint of xⁿ = ²ⁿC_n

The general term for binomial (1 + x)^2n-1 is
T_r+1 = ^2n-1C_r x^r

To find the co-efficient of xⁿ
Putting r = n
T_r+1 = ^2n-1C_r x^r
The coefficient of xⁿ = ^2n-1C_n
We have to prove
Coefficient of xⁿ in (1 + x)²ⁿ = 2 coefficient of xⁿ in (1 + x)^2n-1
Consider LHS = ²ⁿC_n = \(\frac{2 n!}{n!(2 n-n)!}=\frac{2 n!}{n!(n)!}\)
Again consider RHS = 2 × ^2n-1C_n
= 2 × \(\frac{(2 n-1)!}{n!(2 n-1-n)!}=2 \times \frac{(2 n-1)!}{n!(n-1)!}\)
Now multiplying and dividing by n we get, 2 × \(\frac{(2 n-1)!}{n!(n-1)!} \times \frac{n}{n}=\frac{2 n(2 n-1)!}{n!n(n-1)!}=\frac{2 n!}{n!n!}\)
From above equation LHS = RHS.
Hence proved.

Question 8.
Find a positive value of m for which the coefficient of x² in the expansion (1 + x)^m is 6.
Solution:
The general term T_r+1 of the binomial expansion (a + b)ⁿ is given by T_r+1 = ⁿC_r a^n-r b^r
Here, a = 1, b = x and n = m
Putting above values, T_r+1 = ^mC_r 1^n-r x^r = ^mC_r x^r
We need the coefficient of x²
Putting r = 2, T₂₊₁ = ^mC₂ x²
The coefficient of x² = ^mC₂
Given that coefficient of x² = ^mC₂ = 6
⇒ \(\frac{\mathrm{m}!}{2!(\mathrm{m}-2)!}\) = 6
⇒ \(\frac{\mathrm{m}(\mathrm{~m}-1)(\mathrm{m}-2)!}{2 \times 1 \times(\mathrm{m}-2)!}\) = 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 the positive value of m, so m = 4.