Question

Given that the coefficient of the third term in the expansion of (1+3x)^n, in ascending powers of x, is 1539, find the value of n where n is a positive integer ​

Answer 1

Answer:

n = 19

Step-by-step explanation:

$(1 + 3x)^{n} = {1}^{n} + {1}^{n - 1} .(3x)^{1}.nC1 + {1}^{n - 2}.(3x)^{2} .nC2 + ...$

Third term coefficient:

$1^{n - 2}.(3)^{2} .nC2 = 1539 \\ 9.nC2 = 1539 \\ nC2 = 171$

nCr = n!/((n - r)!r!)

nC2 = 171

n!/((n - 2)!2!) = 171

n!/2(n - 2)! = 171

n!/(n - 2)! = 342

n.(n - 1).(n -2).../((n - 2).(n - 3).(n - 4)...) = 342

n(n - 1) = 342

n² - n = 342

(n - ½)² - ¼ = 342

(n - ½)² = 1369/4

(n - ½) = ±37/2

n = 1/2 ± 37/2

n = -18 or n = 19

Since we are told n is a positive integer, it is 19