If _r=0^5^11C_2r+12r+2=(m)/(n),gcd(m,n)=1, then m-n is equal to __________.

Question

If $\sum _{r=0}^{5}\frac{{}^{11}{C}_{2r+1}}{2r+2}=\frac{m}{n},gcd(m,n)=1$, then $m-n$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 2035

Accepted Answer

$\begin{gathered} (1+x{)}^{11}={}^{11}{C}_{0}+{}^{11}{C}_{1}x+{}^{11}{C}_{2}{x}^{2}+\cdots +{}^{11}{C}_{11}{x}^{11} \ {\int }_{1}^{0}(1+x{)}^{11}dx={\int }_{1}^{0}({}^{11}{C}_{0}+{}^{11}{C}_{1}x+{}^{11}{C}_{2}x{}^{2}+\cdots +{}^{11}{C}_{11}x{}^{11})dx \ {{\frac{(1-x{)}^{12}}{12}]}_{1}^{0}={}^{11}{C}_{0x}+\frac{{}^{11}{C}_{1}{x}^{2}}{2}+\frac{{}^{11}{C}_{2}x}{3}+\cdots +\frac{{}^{11}{C}_{11}{x}^{12}}{12}]}_{1}^{0} \end{gathered}$

$\frac{{2}^{12}-1}{12}={C}_{0}+\frac{{C}_{1}}{2}+\frac{{C}_{2}}{3}+\cdots +\frac{{C}_{11}}{12}\cdots$ ........(1)

Now,

$\begin{gathered} {\int }_{0}^{-1}(1+x{)}^{11}dx={\int }_{0}^{-1}({}^{11}{C}_{0}+{}^{11}{C}_{1}x+{}^{11}{C}_{2}{x}^{2}+\cdots +{}^{11}{C}_{11}{x}^{11})dx \ {{\frac{(1+x{)}^{12}}{12}]}_{0}^{-1}={}^{11}{C}_{0}x+\frac{{}^{11}{C}_{1}{x}^{2}}{2}+\frac{{}^{11}{C}_{2}{x}^{3}}{3}+\cdots +\frac{{}^{11}{C}_{11}{x}^{12}}{12}]}_{0}^{-1} \end{gathered}$

$\frac{1}{12}={C}_{0}-\frac{{C}_{1}}{2}+\frac{{C}_{2}}{3}\cdots$ \ldots \ldots .(2)

$\begin{gathered} -(2) \ =\frac{{2}^{12}-2}{12}=2[\frac{{C}_{1}}{2}+\frac{{C}_{3}}{4}+\cdots ] \ \Rightarrow \sum _{r=0}^{5}\frac{{C}_{2r+1}}{2r+2}=\frac{{2}^{11}-1}{12}=\frac{2047}{12}=\frac{m}{n} \ =2047-12=2035 \end{gathered}$

Solutions · 1