12. 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.

1.The general term of the expansion (a + b)n is given byTr +1 = nCran–rbrSo, T1 = nC0an = anT2 = nC1an-1b = n!1!(n−1)! an-1 b = n×(n−1)!(n−1)! an-1b = nan-1bT3 = nC2an-2b2 = n!2!(n−2)[! an-2b2 = n ×(n−1)×(n−2)!2×1×(n−2)! an-2b2 = n(n−1)2 an-2b2Given,T1 = 729=>an = 729 ------------------ (1)T2 = 7290=>nan–1b = 7290 ------------- (2)T3 = 30375=> n(n−1)2 an–2b2 = 30375 ------------------- (3)Dividing equation (2) by (1) we get,nan−1ban = 7290729=> nba = 10Similarly dividing equation (3) by (2) we get,n(n−1)2 an–2b2 ÷ nan–1b = 303757290=> n(n−1)2 an–2b2× 1nan−1b = 303757290=> n(n−1)an−2b2 × 1nan−1b = 303757290 × 2=> (n−1)ba = 25′ ×26=> nba −ba = 253=> 10 – ba = 253 [since, nba = 10]=> ba = 10 – 253= 30−253= 53 --------------- (5)Putting equation (5) in (4) we get,n× 53 = 10=>n = 10 × 35=>n = 6So putting the value of n in equation (1) we get,a6 = 729=>a6= 36=>a = 3And putting a = 3 in equation (5) we get,ba = 53=>b = 53 ×a= 53 × 3= 5

Ask & Answer Question

Binomial Theorem Ncert Solutions Maths class 11th Maths Class 11th Binomial Theorem

12. 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.

2 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

4 months ago

1.The general term of the expansion (a + b)ⁿ is given by

T_r₊₁ = ⁿC_raⁿ^–rb^r

So, T₁ = ⁿC₀aⁿ = aⁿ

T₂ = ⁿC₁aⁿ^-1b = $\frac{n!}{1! (n - 1)!}$ aⁿ^-1 b = $\frac{n \times (n - 1)!}{(n - 1)!}$ aⁿ^-1b = naⁿ^-1b

T₃ = ⁿC₂aⁿ^-2b² = $\frac{n!}{2! (n - 2) [!}$ aⁿ^-2b² = $\frac{n \times (n - 1) \times (n - 2)!}{2 \times 1 \times (n - 2)!}$ aⁿ^-2b² = $\frac{n (n - 1)}{2}$ aⁿ^-2b²

Given,

T₁ = 729

=>aⁿ = 729 ------------------ (1)

T₂ = 7290

=>naⁿ^–1b = 7290 ------------- (2)

T₃ = 30375

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b² = 30375 ------------------- (3)

Dividing equation (2) by (1) we get,

$\frac{n a^{n - 1} b}{a^{n}}$ = $\frac{7290}{729}$

=> $\frac{n b}{a}$ = 10

Similarly dividing equation (3) by (2) we get,

$\frac{n (n - 1)}{2}$ aⁿ^–2b² ÷ naⁿ^–1b = $\frac{30375}{7290}$

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b²× $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$

=> $n (n - 1) a^{n - 2} b^{2}$ × $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$ × 2

=> $\frac{(n - 1) b}{a}$ = $\frac{2 5′ \times 2}{6}$

=> $\frac{n b}{a}$ $- \frac{b}{a}$ =

...more

1.The general term of the expansion (a + b)ⁿ is given by

T_r₊₁ = ⁿC_raⁿ^–rb^r

So, T₁ = ⁿC₀aⁿ = aⁿ

T₂ = ⁿC₁aⁿ^-1b = $\frac{n!}{1! (n - 1)!}$ aⁿ^-1 b = $\frac{n \times (n - 1)!}{(n - 1)!}$ aⁿ^-1b = naⁿ^-1b

T₃ = ⁿC₂aⁿ^-2b² = $\frac{n!}{2! (n - 2) [!}$ aⁿ^-2b² = $\frac{n \times (n - 1) \times (n - 2)!}{2 \times 1 \times (n - 2)!}$ aⁿ^-2b² = $\frac{n (n - 1)}{2}$ aⁿ^-2b²

Given,

T₁ = 729

=>aⁿ = 729 ------------------ (1)

T₂ = 7290

=>naⁿ^–1b = 7290 ------------- (2)

T₃ = 30375

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b² = 30375 ------------------- (3)

Dividing equation (2) by (1) we get,

$\frac{n a^{n - 1} b}{a^{n}}$ = $\frac{7290}{729}$

=> $\frac{n b}{a}$ = 10

Similarly dividing equation (3) by (2) we get,

$\frac{n (n - 1)}{2}$ aⁿ^–2b² ÷ naⁿ^–1b = $\frac{30375}{7290}$

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b²× $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$

=> $n (n - 1) a^{n - 2} b^{2}$ × $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$ × 2

=> $\frac{(n - 1) b}{a}$ = $\frac{2 5′ \times 2}{6}$

=> $\frac{n b}{a}$ $- \frac{b}{a}$ = $\frac{25}{3}$

=> 10 – $\frac{b}{a}$ = $\frac{25}{3}$ [since, $\frac{n b}{a}$ = 10]

=> $\frac{b}{a}$ = 10 – $\frac{25}{3}$

= $\frac{30 - 25}{3}$

= $\frac{5}{3}$ --------------- (5)

Putting equation (5) in (4) we get,

n× $\frac{5}{3}$ = 10

=>n = 10 × $\frac{3}{5}$

=>n = 6

So putting the value of n in equation (1) we get,

a6 = 729

=>a⁶= 36

=>a = 3

And putting a = 3 in equation (5) we get,

$\frac{b}{a}$ = $\frac{5}{3}$

=>b = $\frac{5}{3}$ ×a

= $\frac{5}{3}$ × 3

= 5

less

0 Upvotes 0 Downvotes

Similar Questions for you

If ²⁰C_r is the co-efficient of x^r in the expansion of (1 + x)²⁰, then the value of $\sum_{r = 0}^{20} r^{2}$ ²⁰C_r is equal to

Vishal Baghel

Kindly consider the following figure

If the coefficient of x¹⁰ in the binomial expansion of ${(\frac{\sqrt[]{x}}{5^{\frac{1}{4}}} + \frac{\sqrt[]{5}}{x^{\frac{1}{3}}})}^{60}$ is $5^{k} . l,$ where $l, k \in N a n d l$ is co-prime to 5, then k is equal to………...

alok kumar singh

$T_{r + 1} =^{60} C_{r} {(x^{\frac{1}{2}})}^{60 - r} {(x^{- \frac{1}{3}})}^{r} {(5^{\frac{- 1}{4}})}^{60 - r} {(5^{\frac{1}{2}})}^{r}$

for

$x^{10} \frac{60 - r}{2} - \frac{r}{3} = 10$

$\Rightarrow 180 - 3 r - 2 r = 60$

->r = 24

k = 3 + exponent of 5 in

= $3 + ([\frac{60}{5}] + [\frac{60}{5^{2}}] - [\frac{24}{5}] - [\frac{24}{5^{2}}] - [\frac{36}{5}] - [\frac{36}{5^{2}}])$

= 3 + (12 + 2 – 4 – 0 – 7 – 1)

= 3 + 2 = 5

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

Payal Gupta

15. ${[3 x^{2} - 2 a x + 3 a^{2}]}^{3}$

= ${[3 x^{2} + a (- 2 x + 3 a)]}^{3}$

We know that by binomial theorem,

${(a + b)}^{3}$ = $a^{3} + b^{3} + 3 a b (a + b)$

= $a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}$

Then,

${[3 x^{2} + a (- 2 x + 3 a)]}^{3}$

= (3x²)³ + ${[a (- 2 x + 3 a)]}^{3}$ + $[3 (3 x^{2})^{2} a (- 2 x + 3 a)]$ + $[3 (3 x^{2}) {a (- 2 x + 3 a)}^{2}]$

= 27x⁶ + $[a^{3} {(- 2 x + 3 a)}^{3}]$ + $[3 (9 x^{4}) (- 2 a x + 3 a^{2})]$ + $[3 (3 x^{2}) {a^{2} {(3 a - 2 x)}^{2}}]$

= 27x⁶ + $[a^{3} {- 8 x^{3} + 27 a^{3} + 3 (4 x^{2}) (3 a) + 3 (- 2 x) (9 a^{2})}]$ + $[- 54 a x^{5} + 81 a^{2} x^{4}]$ + [ $(9 a^{2} x^{2})$ $(9 a^{2} + 4 x^{2} - 12 a x)$ ]

= 27x⁶ + [ $- 8 a^{3} x^{3} + 27 a^{6} + 36 a^{4} x^{2} - 54 a^{5} x$ ] + [ $- 54 a x^{5} + 81 a^{2} x^{4}$ ] + [ $(81 a^{4} x^{2} + 36 a^{2} x^{4} - 108 a^{3} x^{3}$ ]

= 27x⁶ $- 8 a^{3} x^{3} + 27 a^{6} + 36 a^{4} x^{2} - 54 a^{5} x$ $- 54 a x^{5} + 81 a^{2} x^{4}$ + $81 a^{4} x^{2} + 36 a^{2} x^{4} - 108 a^{3} x^{3}$

= 27x⁶– 54ax⁵ $+ 117 a^{2} x^{4}$ $- 116 a^{3} x^{3}$ + $117 a^{4} x^{2}$ $- 54 a^{5} x$ $+ 27 a^{6}$

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

Payal Gupta

15. ${[3 x^{2} - 2 a x + 3 a^{2}]}^{3}$

= ${[3 x^{2} + a (- 2 x + 3 a)]}^{3}$

We know that by binomial theorem,

${(a + b)}^{3}$ = $a^{3} + b^{3} + 3 a b (a + b)$

= $a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}$

Then,

${[3 x^{2} + a (- 2 x + 3 a)]}^{3}$

= (3x²)³ + ${[a (- 2 x + 3 a)]}^{3}$ + $[3 (3 x^{2})^{2} a (- 2 x + 3 a)]$ + $[3 (3 x^{2}) {a (- 2 x + 3 a)}^{2}]$

= 27x⁶ + $[a^{3} {(- 2 x + 3 a)}^{3}]$ + $[3 (9 x^{4}) (- 2 a x + 3 a^{2})]$ + $[3 (3 x^{2}) {a^{2} {(3 a - 2 x)}^{2}}]$

= 27x⁶ + $[a^{3} {- 8 x^{3} + 27 a^{3} + 3 (4 x^{2}) (3 a) + 3 (- 2 x) (9 a^{2})}]$ + $[- 54 a x^{5} + 81 a^{2} x^{4}]$ + [ $(9 a^{2} x^{2})$ $(9 a^{2} + 4 x^{2} - 12 a x)$ ]

= 27x⁶ + [ $- 8 a^{3} x^{3} + 27 a^{6} + 36 a^{4} x^{2} - 54 a^{5} x$ ] + [ $- 54 a x^{5} + 81 a^{2} x^{4}$ ] + [ $(81 a^{4} x^{2} + 36 a^{2} x^{4} - 108 a^{3} x^{3}$ ]

= 27x⁶ $- 8 a^{3} x^{3} + 27 a^{6} + 36 a^{4} x^{2} - 54 a^{5} x$ $- 54 a x^{5} + 81 a^{2} x^{4}$ + $81 a^{4} x^{2} + 36 a^{2} x^{4} - 108 a^{3} x^{3}$

= 27x⁶– 54ax⁵ $+ 117 a^{2} x^{4}$ $- 116 a^{3} x^{3}$ + $117 a^{4} x^{2}$ $- 54 a^{5} x$ $+ 27 a^{6}$

14. 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]

Payal Gupta

14. For (a – b) to be a factor of aⁿ – b ⁿwe need to show (aⁿ – bⁿ) = (a – b)k as k is a natural number.

We have, for positive n

aⁿ = ${(a - b + b)}^{n}$ = ${[(a - b) + b]}^{n}$

=>aⁿ = ⁿC₀(a – b)ⁿ + ⁿC₁(a – b)^{n -}¹b + ⁿC₂(a – b)^{n –}²b² + ………… +ⁿC_n_-1 $(a - b) b^{n - 1}$ + ⁿC_nbⁿ

=>aⁿ= ${(a - b)}^{n}$ + ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + …………….…+ ⁿC_n_-1 $(a - b) b^{n - 1}$ + $b^{n}$ [Since, ⁿC₀ = 1 and ⁿC_n = 1]

=> $a^{n} - b^{n}$ = ${(a - b)}^{n}$ +ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + ……………… + ⁿC_n_-1 $(a - b) b^{n - 1}$

=> $a^{n} - b^{n}$ = $(a - b)$ [ ${(a - b)}^{n - 1}$ + ⁿC₁ ${(a - b)}^{n - 2} b$ + ⁿC₂ ${(a - b)}^{n - 3} b^{2}$ +………..…… + ⁿC_n_-1 $b^{n - 1}$ ]

=> $a^{n} - b^{n}$ &n

...more

14. For (a – b) to be a factor of aⁿ – b ⁿwe need to show (aⁿ – bⁿ) = (a – b)k as k is a natural number.

We have, for positive n

aⁿ = ${(a - b + b)}^{n}$ = ${[(a - b) + b]}^{n}$

=>aⁿ = ⁿC₀(a – b)ⁿ + ⁿC₁(a – b)^{n -}¹b + ⁿC₂(a – b)^{n –}²b² + ………… +ⁿC_n_-1 $(a - b) b^{n - 1}$ + ⁿC_nbⁿ

=>aⁿ= ${(a - b)}^{n}$ + ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + …………….…+ ⁿC_n_-1 $(a - b) b^{n - 1}$ + $b^{n}$ [Since, ⁿC₀ = 1 and ⁿC_n = 1]

=> $a^{n} - b^{n}$ = ${(a - b)}^{n}$ +ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + ……………… + ⁿC_n_-1 $(a - b) b^{n - 1}$

=> $a^{n} - b^{n}$ = $(a - b)$ [ ${(a - b)}^{n - 1}$ + ⁿC₁ ${(a - b)}^{n - 2} b$ + ⁿC₂ ${(a - b)}^{n - 3} b^{2}$ +………..…… + ⁿC_n_-1 $b^{n - 1}$ ]

=> $a^{n} - b^{n}$ = $(a - b)$ k where k = [ ${(a - b)}^{n - 1}$ + ⁿC₁ ${(a - b)}^{n - 2} b$ + ⁿC₂ ${(a - b)}^{n - 3} b^{2}$ +………..…… + ⁿC_n_-1 $b^{n - 1}$ ] is a natural number.

Therefore (a – b) is a factor of aⁿ – bⁿwhere n is positive integer.

less

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

12. 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.

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question