Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

# The expression nC0 + 2nC1 + 3nC2 + (n + 1)nCn is equal to

1) (n + 1)2n

2) 2n(n + 2)

3) (n + 2)2n-1

4) (n + 2)2n+1

Solution: (3) (n + 2)2n-1

$$\begin{array}{l}(1+x)^{n}=n_{C_{0}}+n_{C_{1}} x+n_{C_{2}} x^{2}+\cdots+n_{C_{n}} x^{n}\\ \text \ Multiplying \ { }^{\prime} \ x^{\prime} \ on \ both \ sides,\\ x(1+x)^{n}=n_{C_{0}} x+n_{C_{1}} x^{2}+n_{C_{2}} x^{3}+\cdots+n_{c_{n}} x^{n+1}\\ \text \ Differentiating \ with \ respect \ to \ x^{\prime} \ on \ both \ sides,\\ (1+x)^{n}+n x(1+x)^{n-1}= n_{C_{0}}+2 n_{C_{1}} x+3 n_{C_{2}} x^{2}+\cdots +(n+1) n_{c_{n}} x^{n} \\ \text \ Put \ x=1\\ \Rightarrow 2^{\mathrm{n}}+\mathrm{n} 2^{\mathrm{n}-1}=\mathrm{n}_{\mathrm{C}_{0}}+2 \mathrm{n}_{\mathrm{C}_{1}}+3 \mathrm{n}_{\mathrm{C}_{2}}+\cdots+(\mathrm{n}+1) \mathrm{n}_{\mathrm{C}_{\mathrm{n}}}\\ \Rightarrow \mathrm{n}_{\mathrm{C}_{0}}+2 \mathrm{n}_{\mathrm{C}_{1}}+3 \mathrm{n}_{\mathrm{C}_{2}}+\cdots+(\mathrm{n}+1) \mathrm{n}_{\mathrm{C}_{\mathrm{n}}}\\=2^{\mathrm{n}-1}(\mathrm{n}+2)\end{array}$$