If the coefficients of the r-th, (r+1)-th, and (r+2)-th terms in the binomial expansion of…

Question

If the coefficients of the $r$-th, $(r+1)$-th, and $(r+2)$-th terms in the binomial expansion of $(1+y)^m$ are in A.P., then $m$ and $r$ satisfy the equation:

Accepted Answer

Let r = 2

$$\therefore$$ 2nd, 3rd and 4th terms are in AP.

2nd term = T2 = ${}^m{C_1}.y$

Coefficient of T2 = ${}^m{C_1}$

3rd term = T3 = ${}^m{C_2}.{y^2}$

Coefficient of T3 = ${}^m{C_2}$

4th term = T4 = ${}^m{C_3}.{y^3}$

Coefficient of T2 = ${}^m{C_3}$

$$\therefore$$ 2.${}^m{C_2}$ = ${}^m{C_1}$ + ${}^m{C_3}$

$$ \Rightarrow $$ $$2.{{m\left( {m - 1} \right)} \over {1.2}}$$ = $${m \over 1}$$ + $${{m\left( {m - 1} \right)\left( {m - 2} \right)} \over {1.2.3}}$$

$$ \Rightarrow $$ 6m2 - 6m = 6m +m(m2 - 3m + 2)

$$ \Rightarrow $$ 6m2 - 6m = 6m + m3 - 3m2 + 2m

$$ \Rightarrow $$ 6m - 6 = 6 + m2 - 3m + 2

$$ \Rightarrow $$ m2 - 9m + 14 = 0

Now put r = 2 at each option and find answer.

In option C, $${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$$ putting r = 2 we get

m2 - 9m + 14 = 0. So Option C is correct.

Solutions · 1