Exams Joint Entrance Examination Mathematics Binomial Theorem Question 77 Question 77 of 347 Medium NUMERICAL
Let a = 1 + 2 C 2 3 ! + 3 C 2 4 ! + 4 C 2 5 ! + . . . . , b = 1 + 1 C 0 + 1 C 1 1 ! + 2 C 0 + 2 C 1 + 2 C 2 2 ! + 3 C 0 + 3 C 1 + 3 C 2 + 3 C 3 3 ! + . . . . a=1+\frac{{}^{2}{C}_{2}}{3!}+\frac{{}^{3}{C}_{2}}{4!}+\frac{{}^{4}{C}_{2}}{5!}+....,b=1+\frac{{}^{1}{C}_{0}+{}^{1}{C}_{1}}{1!}+\frac{{}^{2}{C}_{0}+{}^{2}{C}_{1}+{}^{2}{C}_{2}}{2!}+\frac{{}^{3}{C}_{0}+{}^{3}{C}_{1}+{}^{3}{C}_{2}+{}^{3}{C}_{3}}{3!}+.... a = 1 + 3 ! 2 C 2 + 4 ! 3 C 2 + 5 ! 4 C 2 + .... , b = 1 + 1 ! 1 C 0 + 1 C 1 + 2 ! 2 C 0 + 2 C 1 + 2 C 2 + 3 ! 3 C 0 + 3 C 1 + 3 C 2 + 3 C 3 + .... Then 2 b a 2 \frac{2b}{{a}^{2}} a 2 2 b is equal to \_\_\_\_\_\_\_\_\_.
Answer: 8
Solutions · 1 Post your solution Official solution
a = 1 + 2 C 2 3 ! + 3 C 2 4 ! + 4 C 2 5 ! + … b = 1 + 1 C 0 + 1 C 1 1 ! + 2 C 0 + 2 C 1 + 2 C 2 2 ! + … b = 1 + 2 1 ! + 2 2 2 ! + 2 3 ! + … = e 2 \begin{gathered} a=1+\frac{{}^{2}{C}_{2}}{3!}+\frac{{}^{3}{C}_{2}}{4!}+\frac{{}^{4}{C}_{2}}{5!}+\ldots \\ b=1+\frac{{}^{1}{C}_{0}+{}^{1}{C}_{1}}{1!}+\frac{{}^{2}{C}_{0}+{}^{2}{C}_{1}+{}^{2}{C}_{2}}{2!}+\ldots \\ b=1+\frac{2}{1!}+\frac{{2}^{2}}{2!}+\frac{2}{3!}+\ldots ={e}^{2} \end{gathered} a = 1 + 3 ! 2 C 2 + 4 ! 3 C 2 + 5 ! 4 C 2 + … b = 1 + 1 ! 1 C 0 + 1 C 1 + 2 ! 2 C 0 + 2 C 1 + 2 C 2 + … b = 1 + 1 ! 2 + 2 ! 2 2 + 3 ! 2 + … = e 2
Using e x = 1 + x 1 ! + x 2 2 ! + x 3 ! + … {e}^{x}=1+\frac{x}{1!}+\frac{{x}^{2}}{2!}+\frac{x}{3!}+\ldots e x = 1 + 1 ! x + 2 ! x 2 + 3 ! x + …
a = 1 + ∑ r = 2 ∞ r C 2 ( r + 1 ) ! = 1 + ∑ r = 2 r ( r − 1 ) 2 ( r + 1 ) ! = 1 + 1 2 ∑ r = 2 ∞ ( r + 1 ) r − 2 r ( r + 1 ) ! = 1 + 1 2 ∑ r = 2 ∞ 1 ( r − 1 ) ! − 1 2 ∑ r = 2 2 r ( r + 1 ) ! = 1 + 1 2 ( 1 1 ! + 1 2 ! + … ) − ∑ r = 2 ∞ ( r + 1 ) − 1 ( r + 1 ) ! = 1 + 1 2 ( e − 1 ) − ∑ r = 2 ∞ 1 r ! + ∑ r = 2 1 ( r + 1 ) ! = 1 + 1 2 ( e − 1 ) − ( e − 1 1 ! − 1 0 ! ) + ( e − 1 1 ! − 1 0 ! − 1 2 ! ) = 1 + e 2 − 1 2 − e + 2 + e − 2 − 1 2 = e 2 ⇒ 2 b a 2 = 2 e 2 4 8 \begin{gathered} a= 1+\sum _{r=2}^{\infty}\frac{{}^{r}{C}_{2}}{(r+1)!}=1+\sum _{r=2}\frac{r(r-1)}{2(r+1)!} \\ =1+\frac{1}{2}\sum _{r=2}^{\infty}\frac{(r+1)r-2r}{(r+1)!} \\ =1+\frac{1}{2}\sum _{r=2}^{\infty}\frac{1}{(r-1)!}-\frac{1}{2}\sum _{r=2}\frac{2r}{(r+1)!} \\ =1+\frac{1}{2}(\frac{1}{1!}+\frac{1}{2!}+\ldots )-\sum _{r=2}^{\infty}\frac{(r+1)-1}{(r+1)!} \\ =1+\frac{1}{2}(e-1)-\sum _{r=2}^{\infty}\frac{1}{r!}+\sum _{r=2}\frac{1}{(r+1)!} \\ =1+\frac{1}{2}(e-1)-(e-\frac{1}{1!}-\frac{1}{0!})+(e-\frac{1}{1!}-\frac{1}{0!}-\frac{1}{2!}) \\ =1+\frac{e}{2}-\frac{1}{2}-e+2+e-2-\frac{1}{2}=\frac{e}{2} \\ \Rightarrow \frac{2b}{{a}^{2}}=\frac{2}{\frac{{e}^{2}}{4}}8 \end{gathered} a = 1 + r = 2 ∑ ∞ ( r + 1 )! r C 2 = 1 + r = 2 ∑ 2 ( r + 1 )! r ( r − 1 ) = 1 + 2 1 r = 2 ∑ ∞ ( r + 1 )! ( r + 1 ) r − 2 r = 1 + 2 1 r = 2 ∑ ∞ ( r − 1 )! 1 − 2 1 r = 2 ∑ ( r + 1 )! 2 r = 1 + 2 1 ( 1 ! 1 + 2 ! 1 + … ) − r = 2 ∑ ∞ ( r + 1 )! ( r + 1 ) − 1 = 1 + 2 1 ( e − 1 ) − r = 2 ∑ ∞ r ! 1 + r = 2 ∑ ( r + 1 )! 1 = 1 + 2 1 ( e − 1 ) − ( e − 1 ! 1 − 0 ! 1 ) + ( e − 1 ! 1 − 0 ! 1 − 2 ! 1 ) = 1 + 2 e − 2 1 − e + 2 + e − 2 − 2 1 = 2 e ⇒ a 2 2 b = 4 e 2 2 8 Answer the question to unlock