( r + 1 ) t h term in the expansion of ( 1 − x ) − 4 will be Options: A . &nbsp x r r ! B . &nbsp ( r + 1 ) ( r + 2 ) ( r + 3 ) 6 x r C . &nbsp ( r + 2 ) ( r + 3 ) 2 x r D . &nbsp None 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

Discussion Forum : Binomial Theorem

Question -

$(r + 1)^{t h}$ term in the expansion of $(1 - x)^{- 4}$ will be

Options:

A .

\frac{x^{r}}{r!}

B .

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

C .

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

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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Discussion Forum : Binomial Theorem

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