$a C_{0} + (a + b) C_{1} + (a + 2 b) C_{2} + . . . . + (a + n b) C_{n}$ is equal to

(2 a + n b) 2^{n}

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

(2 a + n b) 2^{n - 1}

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

(n a + 2 b) 2^{n}

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

(n a + 2 b) 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 $(2 a + n b) 2^{n - 1}$
$\sum_{r = 0}^{n} (a + r b)^{n} c_{r}$

$= \sum_{r = 0}^{n} a \cdot^{n} C_{r} + \sum_{r = 0}^{n} r b^{n} C_{r} \dots (1)$

$Using (1 + x)^{n} =^{n} C_{0} +^{n} C_{1} + \dots +^{n} C_{n}$

$Put x = 1 2^{n} =^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + \dots +^{n} C_{n}$

$\Rightarrow \sum_{r = 0}^{n}^{n} C_{r} = 2^{n}$

(1) $\Rightarrow$

$= a \cdot 2^{n} + b \sum_{r = 0}^{n} r .^{n} c_{r}$

${Using}^{n} C_{r} = \frac{n}{r}^{n - 1} C_{r - 1} = a \cdot 2^{n} + b \sum_{r = 0}^{n} /^{r} \cdot \frac{n}{/^{r}}^{n - 1} C_{r - 1} = a \cdot 2^{n} + b n \cdot 2^{n - 1} = 2^{n - 1} (2 a + n b)$

Suggest Corrections

Similar questions

{(1 + x)}^{n} = n \sum r = 0 C_{r} x^{r}

a C_{0} + (a + b) C_{1} + (a + 2 b) C_{2} + \dots + (a + n b) C_{n}

lim n \to \infty \frac{1^{a} + 2^{a} + . . . . . + n^{a}}{(n + 1)^{a - 1} [(n a + 1) + (n a + 2) + . . . . . + (n a + n)]} = \frac{1}{60}

a

\in R

a \neq - 1, lim n \to \infty \frac{(1^{a} + 2^{a} + \dots + n^{a})}{(n + 1)^{a - 1} [(n a + 1) + (n a + 2) + \dots + (n a + n)]} = \frac{1}{60}

a =

a \in R

a \neq - 1

{lim}_{n \to \infty} \frac{(1^{a} + 2^{a} + . . . . . . + n^{a})}{{(n + 1)}^{a - 1} [(n a + 1) + (n a + 2) + . . . . . . . + (n a + n)]} = \frac{1}{60}

a =

lim n \to \infty \frac{1^{a} + 2^{a} + 3^{a} + . . . + n^{a}}{n^{a + 1}} = \frac{1}{5}

(w h e r e a > - 1)

a

Join BYJU'S Learning Program