Question
Question: Find the value of :- \({C_1} + 2{C_2} + 3{C_3} + ... + n{C_n}\) A.\({2^{n - 1}}\) B.\({2^{n + 1...
Find the value of :- C1+2C2+3C3+...+nCn
A.2n−1
B.2n+1
C.n2n−1
D.n2n+1
Solution
We will use the formula for the expansion of a binomial term and find the binomial expansion of (1+x)n. After simplifying it, we will differentiate both sides of the equation with respect to x. We will use the formula for the differentiation of xn with respect to x to do so. After differentiating the equation, we will substitute 1 for x in the new equation that we have obtained. After substituting 1 in the new equation, we will obtain the given expression on the right-hand side of the equation and the required answer on the left-hand side of the equation. We will choose the correct option accordingly.
Formulas used:
We will use the following formulas:
1.The binomial expansion of (a+b)n is given by (a+b)n=nC0anb0+nC1an−1b1+...+nCna0bn.
2.The differentiation of xn with respect to x is given by dxd(xn)=nxn−1.
Complete step-by-step answer:
We will calculate the binomial expansion of (1+x)n by substituting 1 for a and x for b in the formula (a+b)n=nC0anb0+nC1an−1b1+...+nCna0bn. Therefore, we get
(1+x)n=nC01nx0+nC11n−1x1+nC21n−1x2...+nCn10xn
Simplifying the above equation, we get
⇒(1+x)n=nC0+nC1x1+nC2x2...+nCnxn…………………………………(1)
We will differentiate both sides of equation (1) with respect to x.
⇒dxd(1+x)n=dxd(nC0+nC1x1+nC2x2...+nCnxn)
Using the formula dxd(xn)=nxn−1 x in the above equation, we get
⇒n(1+x)n−1=0+nC1+2nC2x+...+nxn−1
Substituting 1 for x in the above equation, we get