Saved Bookmarks
| 1. |
Ifalpha=(5)/(2!3!) + (5.7)/(3!3^2)+ (5.7.9)/(4!3^3) +…,thenalpha ^ 2 + 4 alphaisequalto |
|
Answer» 21 `alpha= (5 ) /(2!3 )+(5.7 )/(3!3^ 2 )+ (5.7.9)/(4!3^3) +… ` ` (1 + X ) ^n= 1 + (NX )/(1 !)+(n(n - 1 ))/(2!)x ^ 2+(n (n-1) (n - 2 ))/(3!) x ^3` `+ (n(n - 1 )(n - 2 ) (n - 3 ))/(4!)x ^4 + ... "" ` ...(2) On comparingthe termsofabovetwoexpansions, weget, ` (n(n - 1))/(2!)x ^ 2 =(5 )/(2!3)"" `...(1) `(n(n - 1)(n - 2))/(3 ! ) x ^ 3=(5.7 ) /(3 ! 3^ 2 ) "" `...(2) ` (n (n - 1 )(n - 2 )(n - 3 ))/(4 ! )x ^ 4 =(5.7.9)/(4!3^3) "" `...(3) `(2)div (1) ,(3) div (2)` `(n - 2 )x = (7 ) /(3) "" `...(4) ` (n - 3 )x=3 ""`... (5) ` (7)/(3x)- (3 )/(x)=1 "" ` [From(4)and(5) ] `(-2 ) /(3x )=1 ` ` rArrx =(-2)/(3 ) ` `RARR(n - 2 )(( -3 )/(2)) =(7)/(2)` ` rArrn = 2- (7)/(2)` `rArr n =(-3)/(2) ` `therefore(1 -(2)/(3))^(-3/2)= 1 +((-3/2) (-2/3))/(1!)+((-3/2)(-3/2) - (1)/(2!) (-2/3)^2 +... ` [`because` From (2) ] `rArr(1 - (2)/(3)) ^(-3/2) = 1+ 1 + (5 )/(2!3) +(5.7 ) /(3!3^2)+... ` ` rArr3^(3//2)= 2+(5)/(2!3)+(5.7 ) / (3 !3 ^ 2 )+ ... ` `3 ^(3//2)=2+alpha` `rArr (5 ) /2 ! 3 )+(5.7 ) /( 3 ! 3 ^ 2 )+...=3 ` `thereforealpha=3 ^(3//2 ) - 2` `alpha + 2=3 ^(3//2 ) ` ` rArr (alpha+2 ) ^2=3 ^ 2 ` `rArralpha ^ 2+4 alpha=27 - 4` `rArralpha ^ 2+ 4 alpha=23 ` |
|