Question

If for some positive integer n, coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{n+5}$ are in ratio 5 : 10 : 14, then the largest coefficient in this expansion is:

• A. 252
• B. 462
• C. 792
• D. 330

Answer 1

Answer: (B)

462

Answer 2

Let the binomial expansion be $(1+x)^m$, where $m = n+5$. The general term is $T_{r+1} = \binom{m}{r} x^r$. Let the three consecutive terms have indices $k$, $k+1$, and $k+2$. Their coefficients are $\binom{n+5}{k}$, $\binom{n+5}{k+1}$, and $\binom{n+5}{k+2}$. Given ratio: $\binom{n+5}{k} : \binom{n+5}{k+1} : \binom{n+5}{k+2} = 5 : 10 : 14$ This gives two equations:

1. $\frac{\binom{n+5}{k}}{\binom{n+5}{k+1}} = \frac{5}{10} = \frac{1}{2}$
2. $\frac{\binom{n+5}{k+1}}{\binom{n+5}{k+2}} = \frac{10}{14} = \frac{5}{7}$ Using the formula $\frac{\binom{N}{j}}{\binom{N}{j+1}} = \frac{j+1}{N-j}$: For equation 1, with $N=n+5$ and $j=k$: $\frac{k+1}{(n+5)-k} = \frac{1}{2} \implies 2(k+1) = n+5-k \implies 3k = n+3 \quad (*)$ For equation 2, with $N=n+5$ and $j=k+1$: $\frac{(k+1)+1}{(n+5)-(k+1)} = \frac{5}{7} \implies \frac{k+2}{n-k+4} = \frac{5}{7} \implies 7(k+2) = 5(n-k+4) \implies 12k = 5n+6 \quad (**)$ Solving the system of equations $(*)$ and $(**)$: From $(*)$, $n = 3k-3$. Substitute into $(**)$: $12k = 5(3k-3) + 6$ $12k = 15k - 15 + 6$ $12k = 15k - 9 \implies 3k = 9 \implies k = 3$ Substitute $k=3$ back into $n = 3k-3$: $n = 3(3) - 3 = 9 - 3 = 6$ So, $n=6$. The binomial expansion is $(1+x)^{n+5} = (1+x)^{6+5} = (1+x)^{11}$. The largest coefficient in the expansion of $(1+x)^m$ occurs at the middle term(s). For $m=11$, the largest coefficients are $\binom{11}{5}$ and $\binom{11}{6}$. $\binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462$ The largest coefficient is 462.