Misc 1 - Find a, b, n in expansion of (a + b)n if first three - Miscellaneous

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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 b0 + nC1 an – 1 b1 + nC2 an – 2 b2 + …. …. … + nCn – 1 a1 bn – 1 + nCn a0 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 Therefore, an = 729 nC1 an–1 b = 7290 nC2 an–2 b2 = 30375 Dividing (1) by (2) ﷐﷐𝑎﷮𝑛﷯﷮𝑛𝐶1﷐𝑎﷮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﷯﷮𝑛﷐𝑎﷮𝑛−1﷯ 𝑏﷯ = ﷐1﷮10﷯ ﷐﷐𝑎﷮𝑛 −(𝑛−1)﷯﷮𝑛 𝑏﷯ = ﷐1﷮10﷯ ﷐﷐𝑎﷮1﷯﷮𝑛𝑏﷯ = ﷐1﷮10﷯ ﷐𝑎﷮𝑛𝑏﷯ = ﷐1﷮10﷯ Dividing (2) by (3) ﷐𝑛𝐶1﷐𝑎﷮𝑛−1﷯ 𝑏﷮𝑛𝐶2﷐𝑎﷮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

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