For a positive integer n,
( 1 + 1/x )ⁿ is expanded
in increasing powers of x. If three consecutive
coefficients in this expansion are in the ratio,
2 : 5 : 12, then n is equal to________.

Solution

Maths sir · 5 steps

Do not chase the coefficients themselves; divide neighbours, because their ratio gives a neat linear equation.

1. Set up coefficients. Each term is C(n, k)x⁻ᵏ. Writing them in increasing powers of x only reverses the order, and C(n, k) = C(n, n-k), so the coefficients are still binomial coefficients. Take three consecutive ones as C(n, r-1) : C(n, r) : C(n, r+1) = 2 : 5 : 12.

   C(n, r)/C(n, r-1) = (n-r+1)/r

2. First ratio. (n-r+1)/r = 5/2 gives 2n - 2r + 2 = 5r, so 2n + 2 = 7r.
3. Second ratio. C(n, r+1)/C(n, r) = (n-r)/(r+1) = 12/5 gives 5n - 5r = 12r + 12, so 5n = 17r + 12.
4. Solve together. From the first equation, n = (7r-2)/2. Substituting gives 35r - 10 = 34r + 24, so r = 34 and n = (238 - 2)/2 = 118.
5. So. Check: 85/34 = 5/2 and 84/35 = 12/5, so both ratios hold, and n = 118.

The common slip: 34. Stopping after solving for r gives 34, which is the term position, not n. Put it back into 2n + 2 = 7r to get n.

The idea to keep. For consecutive binomial coefficients, always use C(n, r)/C(n, r-1) = (n-r+1)/r. Each given ratio becomes a linear equation in n and r.

Answer from the NTA final answer key. Solution by PrepWiser.