Misc 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.
We know that
(a + b)n = nC0 an + nC1 an – 1 b1 + nC2 an – 2 b2 + ….…. + nCn – 1 a1 bn – 1 + nCn bn
= an + nC1 an – 1 b1 + nC2 an – 2b2 +…. …. + nCn – 1 a1 bn – 1 + bn
So first 3 terms are an , nC1 an – 1b and nC2 an – 2 b2
Also, it is given that their value are 729,7290 and 30375
∴ an = 729
nC1 an – 1 b = 7290
nC2 an – 2 b2 = 30375
Dividing (1) by (2)
𝑎^𝑛/(𝑛𝐶1𝑎^(𝑛 − 1) 𝑏) = 729/7290
𝑎^𝑛/(𝑛!/1!(𝑛−1)! 𝑎^(𝑛 − 1) 𝑏) = 1/10
𝑎^𝑛/((𝑛(𝑛 − 1)!)/1(𝑛 − 1)! 𝑎^(𝑛 − 1) 𝑏) = 1/10
𝑎^𝑛/(𝑛𝑎^(𝑛 − 1) 𝑏) = 1/10
𝑎^(𝑛 − (𝑛 − 1))/𝑛𝑏 = 1/10
𝑎^1/𝑛𝑏 = 1/10
𝑎/𝑛𝑏 = 1/10
Dividing (2) by (3)
(𝑛𝐶1𝑎^(𝑛−1) 𝑏)/(𝑛𝐶2𝑎^(𝑛−2) 𝑏^2 ) = 7290/30375
(𝑛!/(1!(𝑛−1)! ) 〖 𝑎〗^(𝑛−1).𝑏)/(𝑛!/2!(𝑛 −2)! 𝑎^(𝑛−2) 𝑏2) = 6/25
(𝑛!〖 𝑎〗^(𝑛 − 1) 𝑏)/(𝑛 − 1)! × 2!(𝑛 − 2)!/(𝑛!𝑎^(𝑛 − 2) 𝑏2) = 6/25
(𝑛! 2! (𝑛 − 2)!)/𝑛!(𝑛 − 1)! × 𝑎^(𝑛 − 1)/𝑎^(𝑛 − 2) × 𝑏/𝑏2 = 6/25
(2 𝑛! (𝑛 − 2)!)/𝑛!(𝑛 − 1)(𝑛 − 2)! × 𝑎^(𝑛−1)/𝑎^(𝑛−2) × 𝑏/𝑏2 = 6/25
2/((𝑛 − 1) ) × 𝑎^(𝑛 −1 − (𝑛 − 2))/1 × 𝑏/𝑏2 = 6/25
2/((𝑛 − 1)) × 𝑎^(𝑛 − 1 − 𝑛 + 2)/𝑏 = 6/25
2𝑎/(𝑛 − 1)𝑏 = 6/25
Now, our equations are
𝑎/𝑛𝑏 = 1/10 …(4)
2𝑎/(𝑛−1)𝑏 = 6/25 …(5)
Dividing (4) by (5)
(𝑎/𝑛𝑏)/(2𝑎/(𝑛−1)𝑏) = (1/10)/(6/25)
𝑎/𝑛𝑏 × (𝑛 − 1)𝑏/2𝑎 = 1/10 × 25/6
((𝑛 − 1))/2𝑛 = 5/12
12 (n – 1) = 2n (5)
12n – 12 = 10n
12n – 10n = 12
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)
2𝑎/(𝑛−1)𝑏 = 6/25
3/(6 × 𝑏) = 1/10
1/2𝑏 = 1/10
10/2 = b
5 = b
b = 5
Hence,
a = 3 , b = 5 & n = 6

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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