Find the value of $^{n} C_{1} + 2^{n} C_{2} + 3^{n} C_{3} . . . . . . . . . n^{n} C_{n}$

$n \cdot 2^{n}$

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

$(n - 1)^{2 n}$

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!

$n \cdot 2^{n - 1}$

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

Open in App

Solution

The correct option is D
$n \cdot 2^{n - 1}$

Method 1(By differentiation)
${(1 + x)}^{n} = n C 0 + n C 1 x + n C 2 x^{2} + . . . . + n C n x^{n}$
Differentiate both the sides
$n {(1 + x)}^{n - 1} = n C 1 + 2 n C 2 x + 3 n C 3 x^{2} + . . . . + n n C n x^{n - 1}$

⇒ put x = 1

$n 2^{n - 1} = n C 1 + 2 n C 2 + 3 n C 3 + . . . . + n n C n$

Method 2

Each term of the sum is r $^{n} C_{r}$ , where r varies from 1 to n

$^{n} C_{r} = \frac{n}{r}$ $^{n - 1} C_{r - 1}$

⇒ $r^{n} C_{r} = n^{n - 1} C_{r - 1}$

$n C 1 + 2 n C 2 + 3 n C 3 + . . . . + n n C n$ $= n \sum r = 1 r \cdot^{n} C_{r}$

= $n \sum r = 1 n \cdot^{(n - 1)} C_{(r - 1)}$

$= n \cdot n \sum r = 1$ $^{(n - 1)} C_{(r - 1)}$

= $n \cdot [^{n - 1} C_{0}$ + $^{n - 1} C_{1}$ ..................... $^{n - 1} C_{n - 1}]$

= $n 2^{n - 1}$

[Because $(1 + x)^{n - 1}$ = $^{n - 1} C_{0}$ + $^{n - 1} C_{1} x$ .......... $^{n - 1} C_{n - 1}$ $x^{n - 1}$
If we put $x = 1$ , we get
$\Rightarrow 2^{n - 1} =^{n - 1} C_{0}$ + $^{n - 1} C_{1}$ + ............. $^{n - 1} C_{n - 1}$

Suggest Corrections