Question

In the expansion of (1+x)^29 , the coefficient of (r+1)th term is equal to that of (r+k)th term, then the value of k cannot be

Answer 1

Answer:

The value of k depends on the value of r

Step-by-step explanation:

Given as :

In the expansion of $(1 + x)^{29}$ , the coefficient of ( r+1 ) th term is equal to that of ( r+k ) th term .

Now,

For expansion of $(1 + x)^{n}$ ,

$t_n_+_1$ = $_{r}^{n}\textrm{C}$ × $(x)^{r}$

Similarly ,

For expansion of $(1 + x)^{29}$

$t_r_+_1$ = $_{r}^{29}\textrm{C}$ × $(x)^{r}$

or, $t_r$ = $_{r-1}^{29}\textrm{C}$ × $(x)^{r - 1}$

So, Co-efficient of $t_r$ = $_{r-1}^{29}\textrm{C}$

Now, replace r by r + k

So, $t_r_+_k$ = $_{r+k-1}^{29}\textrm{C}$ × $(x)^{r+k - 1}$

According to question

( r + 1 ) th term = ( r + k ) th term

So,  $_{r-1}^{29}\textrm{C}$ = $_{r+k-1}^{29}\textrm{C}$

Now, From the property

$_{a}^{n}\textrm{C}$  = $_{b}^{n}\textrm{C}$

So, a + b = n

∴  ( r - 1 ) + ( r + k + 1 ) = 29

Or, 2 r + k = 29

So, k = 29 - 2 r

Hence, The value of k depends on the value of r Answer