$C_{1} + 2.5 C_{2} + {3.5}^{2} C_{3} + - - - =$

n {.6}^{n - 1}

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

6^{n - 1}

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

6^{n}

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

n {.6}^{n}

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

Open in App

Solution

The correct option is A $n {.6}^{n - 1}$
$(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} +^{n} C_{3} x^{3} + . . . . .$
$n (1 + x)^{n - 1} =^{n} C_{1} + 2^{n} C_{2} x + 3^{n} C_{3} x^{2} + . . . .$
Put $x = 5$
$n (1 + 5)^{n - 1} = C_{1} + 2.5 C_{2} + {3.5}^{2} C_{3} + . . . .$
$= n 6^{n - 1}$ .

Suggest Corrections

Similar questions

n

(1 + i)^{6 n} + (1 - i)^{6 n} =

^{'} n^{'}

(1 + i)^{6 n} + (1 - i)^{6 n}

6, 6^{2}, 6^{3}, 6^{4} \dots 6^{n}

{(\sqrt{6})}^{n + 1}

\frac{6^{n + 3} - 32 \times 6^{n + 1}}{6^{n + 2} - 2 \times 6^{n + 1}}

^{n} C_{1} + 2 \cdot 5^{n} C_{2} + 3 \cdot 5^{2}^{n} C_{3} + \dots

n

Join BYJU'S Learning Program