If in the expansion of $(1 + x)^{20}$ ,the coefficients of rth and (r+4)th terms are equal,then r is equal to

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Solution

The correct option is C 9
From given, we have,

$(1 + x)^{20}$

and $r$ th term is equal to $(r + 4)$ th term

$\Rightarrow a = 1, b = x, n = 20$

$T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}$

$\Rightarrow T_{r + 1} =^{20} C_{r} 1^{n - r} x^{r}$

$∴ T_{r + 1} =^{20} C_{r} x^{r}$

put $r = r - 1$

$∴ T_{r} =^{20} C_{r - 1} x^{r - 1}$ ..........(1)

put $r = r + 3$

$∴ T_{r + 4} =^{20} C_{r + 3} x^{r + 3}$ .......(2)

now,

$^{20} C_{r - 1} =^{20} C_{r + 3}$

we have formula,

$^{n} C_{r} =^{n} C_{p} \to r = n - p$

$\Rightarrow r - 1 = 20 - (r + 3)$

$r - 1 = 20 - r - 3$

$2 r = 18$

$∴ r = 9$

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