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Question:

If three consecutive coefficients in the expansion of (x + y)ⁿ are in the ratio 1 : 9 : 63, then the value of n is

Explanation:

We know (x + y)ⁿ = ⁿC₀ × xⁿ + ⁿC₁ × x^n-1 × y¹ + ⁿC₂ × x^n-2 × y² + ... + ⁿC_n × yⁿ

Let the three consecutive coefficients be ⁿC_r-1, ⁿC_r & ⁿC_r+1

Now, ⁿC_r-1 : ⁿC_r = 1 : 9
⇒ $\frac{\mathrm{n}!}{(\mathrm{r}-1)!\times (\mathrm{n}-\mathrm{r}+1)!}$ : $\frac{\mathrm{n}!}{\left(\mathrm{r}\right)!\times (\mathrm{n}-\mathrm{r})!}$ = 1 : 9

⇒ $\frac{\mathrm{r}}{\mathrm{n}-\mathrm{r}+1}$ = $\frac{1}{9}$

⇒ 9r = n - r + 1

⇒ 10r = n + 1   ...(1)

Also, ⁿC_r : ⁿC_r+1 = 9 : 63 = 1 : 7
⇒ $\frac{\mathrm{n}!}{\left(\mathrm{r}\right)!\times (\mathrm{n}-\mathrm{r})!}$ : $\frac{\mathrm{n}!}{(\mathrm{r}+1)!\times (\mathrm{n}-\mathrm{r}-1)!}$ = 1 : 7

⇒ $\frac{\mathrm{r}+1}{\mathrm{n}-\mathrm{r}}$ = $\frac{1}{7}$

⇒ 7r + 7 = n - r

⇒ 8r = n - 7   ...(2)

Solving (1) & (2), we get

n = 39 & r = 4

Hence, 39.