Get Answers to all your Questions

header-bg qa

C_1+2 C_2+3 C_3+\ldots .n C_n=

Option: 1


Option: 2

\text { n. } 2^n

Option: 3

\text { n. } 2^{n-1}

Option: 4

n \cdot 2^{n+1}

Answers (1)


We know that, (1+x)^n=C_0+C_1 x+C_2 x^2+\ldots \ldots+C_n x^n

Differentiating both sides w.r.t. x, we get n(1+x)^{n-1}=0+C_1+2 \cdot C_2 x+3 C_3 x^2+\ldots \ldots+n C_n x^{n-1}

Putting x=1 \text {, we get, } n \cdot 2^{n-1}=C_1+2 C_2+3 C_3+\ldots \ldots+n C_n

Posted by

Ritika Kankaria

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE