Question

If the coefficient of $r^{th}$, $(r+1)^{th}$ and $(r+2)^{nd}$ terms in the expansion of $(1+x)^{n}$ are in A.P., then show that $n^{2}-(4r+1)n+4r^{2}-2=0$.

Answer 1

Co-efficient of (r)th term in the expansion of (1 + x)ⁿ is nC(r-1)

Co-efficient of (r + 1)th term in the expansion of (1 + x)ⁿ is nC(r)

Co-efficient of (r + 2)th term in the expansion of (1 + x)ⁿ is nC(r+1)

Given that the above terms are in AP :

⇒ 2nCr = nC(r-1) + nC(r+1)

⇒ $\frac{2}{(n-r)! (r)!}$ = $\frac{1}{[n - (r-1)]! (r-1)!}$ + $\frac{1}{[n-(r+1)]! (r+1)!}$

⇒ $\frac{2(r+1)}{(n-r)! (r+1)!}$ = $\frac{r(r+1)}{[n-(r-1)]! (r+1)!}$ + $\frac{1}{[n-(r+1)]! (r+1)!}$

⇒ $\frac{2(r+1)}{(n-r) [(n-(r+1)]!}$ = $\frac{r(r+1)}{[n-(r-1)] (n-r) [(n-(r+1)]! }$ + $\frac{1}{[n-(r+1)]!}$

⇒ $\frac{2(r+1)}{(n-r)}$ = $\frac{r(r+1)}{[n-(r-1)] (n-r)}$ + 1

⇒ 2(r + 1) × [(n - (r - 1)] = r(r+1) + [n - (r - 1)] × (n-r)

⇒ 2(nr + n - r² + 1) = r² + r + [n² - nr - (rn - r² - n + r)]

⇒ 2nr + 2n - 2r² + 2 = r² + r + n² - rn - rn + r² + n - r  

⇒ 2nr + 2n - 2r² + 2 = 2r² - 2nr + n² + n

⇒ n² + 4r² - 4nr - n - 2 = 0

⇒ n² - n(4r + 1) + 4r² - 2 = 0

Answer 2

Answer:

Co-efficient of (r)th term in the expansion of (1 + x)ⁿ is nC(r-1)

Co-efficient of (r + 1)th term in the expansion of (1 + x)ⁿ is nC(r)

Co-efficient of (r + 2)th term in the expansion of (1 + x)ⁿ is nC(r+1)

Given that the above terms are in AP :

⇒ 2nCr = nC(r-1) + nC(r+1)

⇒ \frac{2}{(n-r)! (r)!} (n−r)!(r)!2 = \frac{1}{[n - (r-1)]! (r-1)!} [n−(r−1)]!(r−1)!1 + \frac{1}{[n-(r+1)]! (r+1)!} [n−(r+1)]!(r+1)!1

⇒ \frac{2(r+1)}{(n-r)! (r+1)!} (n−r)!(r+1)!2(r+1) = \frac{r(r+1)}{[n-(r-1)]! (r+1)!} [n−(r−1)]!(r+1)!r(r+1) + \frac{1}{[n-(r+1)]! (r+1)!} [n−(r+1)]!(r+1)!1

⇒ \frac{2(r+1)}{(n-r) [(n-(r+1)]!} (n−r)[(n−(r+1)]!2(r+1) = \frac{r(r+1)}{[n-(r-1)] (n-r) [(n-(r+1)]! } [n−(r−1)](n−r)[(n−(r+1)]!r(r+1) + \frac{1}{[n-(r+1)]!} [n−(r+1)]!1

⇒ \frac{2(r+1)}{(n-r)} (n−r)2(r+1) = \frac{r(r+1)}{[n-(r-1)] (n-r)} [n−(r−1)](n−r)r(r+1) + 1

⇒ 2(r + 1) × [(n - (r - 1)] = r(r+1) + [n - (r - 1)] × (n-r)

⇒ 2(nr + n - r² + 1) = r² + r + [n² - nr - (rn - r² - n + r)]

⇒ 2nr + 2n - 2r² + 2 = r² + r + n² - rn - rn + r² + n - r

⇒ 2nr + 2n - 2r² + 2 = 2r² - 2nr + n² + n

⇒ n² + 4r² - 4nr - n - 2 = 0

⇒ n² - n(4r + 1) + 4r² - 2 = 0

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