Question

$\frac{{}^n{C}_0}{n}+\frac{{}^n{C}_1}{n+1}+\frac{{}^n{C}_2}{n+2}+\dots \dots +\frac{{}^n{C}_n}{2n}=$

• A. ${\int }_0^1{x}^n(x-1{)}^{n-1}dx$
• B. ${\int }_1^2{x}^n(x-1{)}^{n-1}dx$
• C. ${\int }_0^1{x}^{n-1}(1+x{)}^{n}dx$
• D. ${\int }_1^2{x}^{n-1}(1+x{)}^{n}dx$

Answer 1

Answer: (B), (C)

The correct options are
B ${\int }_1^2{x}^n(x-1{)}^{n-1}dx$
C ${\int }_0^1{x}^{n-1}(1+x{)}^{n}dx$

Given sum ${=}^n{C}_0{\int }_0^1{x}^{n-1}{+}^n{C}_1{\int }_0^1{x}^{n}dx+\dots \dots {+}^n{C}_n{\int }_0^1{x}^{2n-1}dx$
$={\int }_0^1{[}^n{C}_0.x^{n-1}{+}^n{C}_1.x^n+\dots \dots {+}^n{C}_n.x^{2n-1}]dx$
$={\int }_0^1{x}^{n-1}(1+x{)}^{n}dx={\int }_1^2(x-1{)}^{n-1}.x^{n}dx$