Breath Math

1. Find the coefficient of x⁵ in (x + 3)⁸

Solution:

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 x⁵ occurs in the (r + 1)th term of the expansion (x + 3)⁸, we obtain

T_r+1  = ⁸C_r (x)^8-r(3)^r

Comparing the indices of x in x⁵ and in T_{r +1},

we obtain r = 3

Thus, the coefficient of x⁵ is ⁸C₃(3)³ = ^8!/_3!5! x 3³ = ^8.7.6.5!/_3.2.5! .3³ = 1512

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2. Find the coefficient of a⁵b⁷ in (a – 2b)¹²

Solution:

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 a⁵b⁷ occurs in the (r + 1)^th term of the expansion (a – 2b)¹², obtain

T_r+1 = ¹²C_r(a)^12-r(-2b)^r = ¹²C_r (-2)^r(a)^12-r(b)^r

Comparing the indices of a and b in a⁵b⁷ and in T_r+1

we obtain r = 7

Thus the coefficient of a⁵b⁷ is

¹²C₇(-2)⁷ = –^12!/_7!5! .2⁷ = –^{12.11.10.9.8.7!}/_5.4.3.2.7! .2⁷ = -(792)(128) = -101376

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3. Write the general term in the expansion of (x² – y)⁶

Solution:

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

Thus the general term in the expansion of (x² – y)⁶ is

T_r+1 = ⁶C_r(x²)^6-r(-y)^r = (-1)^{r 6}C_r (x)^12-2r(y)^r

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4. Write the general term in the expansion of (x² – yx)¹², x≠0

Solution:

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

Thus the general term in the expansion of (x² – yx)¹²is

T_r+1 = ¹²C_r(x²)^12-r(-yx)^r = (-1)^{r 12}C_r (x)^24-r(y)^r

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

Solution:

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

Thus, the 4^th term in the expansion of (x – 2y)¹²

T₄ = T₃₊₁ = ¹²C₃ (x)^12-3(-2y)³ = (-1)^{3 12!}/_3!9! .x⁹.(2)³.y³ = – ^12.11.10/_3.2 .(2)³.x⁹y³ = -1760x⁹y³

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6. Find the 13^th term in the expansion of (9x – ¹/_3√x)¹⁸, x≠0

Solution:

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

The 13^th term in the expansion of (9x – ¹/_3√x)¹⁸, x≠0

T₁₃ = T₁₂₊₁ = ¹⁸C₁₂ (9x)^18-12 (– ¹/_3√x)¹²

=(-1)¹² . ^18!/_12!6! (9)⁶(x)⁶ (¹/₃)¹²(¹/_√x)¹²

= ^{18.17.16.15.14.13.12!}/_{12!6.5.4.3.2.} x⁶.(¹/_x⁶).3¹².(¹/₃¹²)

= 18564

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7. Find the middle terms in the expansion of [3 – (x³/₆)]⁷

Solution:

It is known that in the expansion of (a+b)ⁿ, if n is odd, then there are two middle terms, namely (ⁿ⁺¹/₂)^th term and (ⁿ⁺¹/₂ + 1)^th term.

Therefore the middle terms in the expansion [3 – (x³/₆)]⁷ are (⁷⁺¹/₂)^th = 4^th term and (⁷⁺¹/₂ + 1)^th = 5^th term

T₄ = T₃₊₁ = ⁷C₃ (3)^7-3 [-(x³/₆)]³ = (-1)^{3 7!}/_3!4! . 3⁴.(x⁹/₆³)

= –^7.6.5.4!/_3.2.4! . 3⁴.[¹/₂³.₃³].x⁹ = –¹⁰⁵/₈ .x⁹

T₅ = T₄₊₁ = ⁷C₄ (3)^7-4 [-(x³/₆)]⁴ = (-1)^{4 7!}/_3!4! . 3³.(x¹²/₆⁴)

= –^7.6.5.4!/_3.2.4! . [(3³)/₂⁴.₃⁴].x¹² = ³⁵/₄₈ .x¹²

Thus, the middle term in the expansion of [3 – (x³/₆)]⁷ are –¹⁰⁵/₈.x⁹ and ³⁵/₄₈.x¹²

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8. Find the middle term in the expansion of (^x/₃ + 9y)¹⁰

Solution:

It is known that in the expansion (a+b)ⁿ, if n is even, then the middle term is (ⁿ/₂ + 1)^th term.

Therefore the middle term in the expansion of (^x/₃ + 9y)¹⁰ is (¹⁰/₂ + 1)^th term

Therefore, the middle term in the expansion of (^x/₃ + 9y)¹⁰ is (¹⁰/₂ + 1)^th = 6^th

T₆ = T₅₊₁ = ¹⁰C₅ (^x/₃)^10-5 (9y)⁵ =^10!/_5!5! . 9⁵.(x⁵/₃⁵).y⁵

= –^{10.9.8.7.6.5!}/_5.4.3.2.5! . [¹/₃⁵].3¹⁰.x⁵.y⁵

=252.3⁵.x⁵.y⁵

Thus, the middle term in the expansion of (^x/₃ + 9y)¹⁰ is 61236x⁵y⁵

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9. In the expansion of (1 + a)^m+n prove that coefficient of a^m and aⁿ are equal

Solution:

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, (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 = ^m+nC_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 th indices of a and aⁿ and in T_k+1

we obtain

k = n

herefore the coefficient of aⁿ is

^m+nC_n = ^(m+n)!/_n!(m+n-m)! =^(m+n)!/_n!m!  ——————-(2)

Thus from (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.

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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.

Solution:

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_r 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 n}C_{r -2}, ⁿC_r-1 ^{and n}C_r 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

ⁿC_r-2/ⁿC_r-1  = ^n!/_{(r-2)!(n-r+2)!} x ^{(r-1)(n-r+1)!}/_n! = ^{(r-1)(r-2)(n-r+1)!}/ ^!/_{(r-2)!(n-r+2)!(n-r+1)!}

= ^r-1/_n-r+2

^r-1/_n-r+2 = ¹/₃

3r – 3 = n – r + 2

n – 4r + 5 = 0

ⁿC_r-1/ⁿC_r  = ^n!/_{(r-1)!(n-r+1)!} x ^(r!)(n-r)!/_n! = ^{r(r-1)!(n-r)!}/_{(r-1)!(n-r+1)(n-r)!}

=^r/_n-r+1

^r/_n-r+1 = ³/₅

5r = 3n – 3r + 3

3n – 8r + 3 = 0

Multiplying (1) by 3 and subtracting it from (2), we obtain

4r – 12 = 0

⇒ r = 3

Putting the value of r in (1), we obtain n

– 12 + 5 = 0

⇒ n = 7

Thus, n = 7 and r = 3

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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

Solution:

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 = ²ⁿC_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_r+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! = ¹/₂[^(2n)!/_(n!)²] ————(2)

From (1) and (2) it is observed that

¹/₂ (²ⁿC_n) =^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.

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

Solution:

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 = ²ⁿC_r a^n-r b^r

Assuming that x² occurs in the (r + 1)^th term of the expansion of (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 in the expansion of (1 + x)^m is 6.

^mC₂ = 6

^(m)!/_2!(m-2)! =6

^{(m)(m-1)(m-2)!}/_2x(m-2)! = 6

m(m – 1) = 12

m² – m – 12 = 0

m² – 4m + 3m – 12 = 0

m(m-4)+3(m – 4) = 0

(m – 4)(m + 3) = 0

(m – 4) = 0 or (m + 3) = 0

m = 4 or m = -3

Thus, the positive value of m for which the coefficient of x² in the expansion (1+ x)^m is 6, is 4.

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