Question

In the binomial expansion (x+y) ^18, find the value of T8/T7

Answer 1

In the binomial expansion (x+y) ^18, the value of T8/T7 is 12y/7x

Given,

(x+y) ^18

The binomial expansion of (x+y) ^18 is given by

T_{r+1} = 18Cr × x^{18-r} × y ^r

To find,

T_8 and T_7

T_8 = T_{7+1} = 18C7 × x^{18-7} × y^7

∴ T_8  = 31824 × x^11 × y ^7

T_7 = T_{6+1} = 18C6 × x^{18-6} × y^6

∴ T_7  = 18564 × x^12 × y^6

T_8 / T_7 = 31824 × x^11 × y^7 / 18564 × x^12 × y^6

∴ T_8 / T_7 = 12y/7x

Answer 2

   $\mathbf{\frac{T_{8}}{T_{7}}= \frac{ 12 \ x}{7\ y}}$

Step-by-step explanation:

• We know some fundamental relation

        $\mathbf{_{r}^{n}\textrm{C}=\frac{!n}{!r\times !(n-r)}}$

        We also know

        !(n-r) = (n-r) !(n-r-1)       Where n is natural number

• In binomial expansion of $\mathbf{(x+y)^{n}}$ , $r^{th}$ term will be

        $\mathbf{T_{r}=_{r-1}^{n}\textrm{C} \ x^{r-1} \ y^{n-r+1}}$

• So In binomial expansion of $\mathbf{(x+y)^{18}}$ , $8^{th}$ and $7^{th}$  term will be

        $\mathbf{T_{8}=_{8-1}^{18}\textrm{C} \ x^{8-1} \ y^{18-8+1}}$

         so

         $\mathbf{T_{8}=_{7}^{18}\textrm{C} \ x^{7} \ y^{11}}$         ...1)

• $\mathbf{T_{7}=_{7-1}^{18}\textrm{C} \ x^{7-1} \ y^{18-7+1}}$

        so

        $\mathbf{T_{7}=_{6}^{18}\textrm{C} \ x^{6} \ y^{12}}$          ...2)

• Taking ratio of equation 1) and equation 2), we get

         $\mathbf{\frac{T_{8}}{T_{7}}=\frac{_{7}^{18}\textrm{C} \ x^{7} \ y^{11}}{_{6}^{18}\textrm{C} \ x^{6} \ y^{12}}}$

         On cancel out x and y , we get

         $\mathbf{\frac{T_{8}}{T_{7}}=\frac{_{7}^{18}\textrm{C} \ x}{_{6}^{18}\textrm{C} \ y}}$

• $\mathbf{\frac{T_{8}}{T_{7}}= \frac{\frac{!18}{!7\times !11} \ x}{\frac{!18}{!6\times !12} \ y}}$

• Above equation can be written as on cancel out !18 in numerator and denominator.

        $\mathbf{\frac{T_{8}}{T_{7}}= \frac{!6\times !12 \ x}{!7\times !11\ y}}$

        Or

        $\mathbf{\frac{T_{8}}{T_{7}}= \frac{!6\times 12\times !11 \ x}{7\times !6\times !11\ y}}$

        Means

        $\mathbf{\frac{T_{8}}{T_{7}}= \frac{ 12 \ x}{7\ y}}$  This is answer.