lf $a_{r}$ is the coefficient of $x^{r}$ in the expansion of $(1 - 2 x + 3 x^{2})^{n}$ then $2 n \sum r = 0 r a_{r} =$

n {.2}^{n}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n {.2}^{n + 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(n + 1) {.2}^{n}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(n + 1) {.2}^{n + 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $n {.2}^{n + 1}$
$\sum_{r = 0}^{r = 2 n} r a_{r}$
$= {\frac{d (1 - 2 x + 3 x^{2})^{n}}{d x}}_{x = 1}$
$= [n (1 - 2 x + 3 x^{2})^{n - 1} (6 x - 2)]_{x = 1}$
$= n (2)^{n - 1} \times 4$
$= n 2^{n + 1}$

Suggest Corrections

Similar questions

x^{r} [0 \leq r \leq (n - 1)]

{(x + 3)}^{n - 1} + {(x + 3)}^{n - 2} (x + 2) + {(x + 3)}^{n - 3} {(x + 2)}^{2} + . . . . . + {(x + 2)}^{n - 1}

\frac{1}{x}

(1 + x)^{n} {(1 + \frac{1}{x})}^{n}

\frac{1}{x}

{(1 + x)}^{n} {(1 + \frac{1}{x})}^{n}

\frac{n!}{[(n - 1)! (n + 1)!]}

\frac{(2 n)!}{[(n - 1)! (n + 1)!]}

\frac{(2 n)!}{(2 n - 1)! (2 n + 1)!}

\frac{2}{1} . \frac{2}{3} . \frac{4}{3} . \frac{4}{5} . \frac{6}{5} . . . . \frac{2 n - 2}{2 n - 3} . \frac{2 n - 2}{2 n - 1} . \frac{2 n}{2 n - 1}

n

x^{r} (0 \geq r \geq n - 1)

(x + 2)^{n - 1} + (x + 2)^{n - 2} (x + 1) + (x + 2)^{n - 3} (x + 1)^{2} + . . . . . + (x + 1)^{n - 1}

^{n} C_{r} (2^{n - r} - 1)

Join BYJU'S Learning Program