The sum of the series ${(1.2)}^{n} C_{2} + {(2.3)}^{n} C_{3} + . . . + {((n - 1) . n)}^{n} C_{n}$ is

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

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

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

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

n (n - 1) 2^{n}

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

None of these

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

Open in App

Solution

The correct option is C $n (n - 1) 2^{n - 2}$
We have,
${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}$ ...(1)
Differentiating equation (1) w.r.t. $x,$ we get
$n {(1 + x)}^{2} = C_{1} + 2 C_{2} x + 3 C_{3} x^{2} + . . . + n C_{n} x^{n - 1}$ ...(2)
Differentiating equation (2) w.r.t. $x,$ we get
$n (n - 1) {(1 + x)}^{n - 2} = (1.2) C_{2} + (2.3) C_{2} x + . . . + n (n - 1) C_{n} x^{n - 2}$ ...(3)
Putting $x = 1$ in equation (3), we have
$(1.2) C_{2} + (2.3) C_{3} + . . . + (n - 1) n C_{n} = n (n - 1) 2^{n - 2} .$

Suggest Corrections

Similar questions

C_{0}, C_{1}, C_{2}, \dots, C_{n}

(1 + x)^{n},

1^{2} \cdot C_{1} + 2^{2} \cdot C_{2} + 3^{2} \cdot C_{3} + \dots + n^{2} C_{n}

\frac{1}{\log_{2} 4} + \frac{1}{\log_{4} 4} + \frac{1}{\log_{8} 4} + . . . . + \frac{1}{{\log_{2}}^{n} 4}

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

\frac{n (n + 1) (2 n + 1)}{12}

\frac{n (n + 1)}{4}

n \geq 2

^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + \dots +^{n} C_{2})

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

With usual notations, $C_{1} + 2^{2} C_{2} + 3^{2} C_{3} + . . . . . . . =$

Join BYJU'S Learning Program