( r + 1 ) t h term in the expansion of ( 1 − x ) − 4 will be Options: A . &nbspxrr! B . &nbsp(r+1)(r+2)(r+3)6xr C . &nbsp(r+2)(r+3)2xr D . &nbspNone of these

Answer: Option B : B ( 1 − x ) − 4 = [ 1.2.3 6 x 0 + 2.3.4 6 x + 3.4.5 6 x 2 + 4.5.6 6 x 3 + . . . . . . . + ( r + 1 ) ( r + 2 ) ( r + 3 ) 6 x r + . . . . . . . . . ] Therefore T r + 1 = ( r + 1 ) ( r + 2 ) ( r + 3 ) 6 x r .

(r+1)th term in the expansion of (1−x)−4 will be

Question

(r + 1)^{t h}

term in the expansion of

(1 - x)^{- 4}

will be

Options:

A . xrr!

B . (r+1)(r+2)(r+3)6xr

C . (r+2)(r+3)2xr

D . None of these

Answer: Option B
:
B

(1 - x)^{- 4}

= [

\frac{1.2.3}{6} x^{0} + \frac{2.3.4}{6} x + \frac{3.4.5}{6} x^{2} + \frac{4.5.6}{6} x^{3} + . . . . . . . + \frac{(r + 1) (r + 2) (r + 3)}{6} x^{r} + . . . . . . . . .]

Therefore

T_{r + 1}

\frac{(r + 1) (r + 2) (r + 3)}{6} x^{r}

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