Quantitative Aptitude For Competitive Examinations (English) 7th Edition

Given (x+1/x) then find value of (x

^{2}+1/x^{2}) (x^{3}+1/x^{3}) (x^{4}+1/x^{4}) …..
Let’s say (x+1/x) =4

Then Find value of (x

^{2}+1/x^{2})
(x

^{2}+1/x^{2}) can be written as (x+1/x)^{2}-2
(x

^{2}+1/x^{2})= 4*4 -2 = 14
Let’s say of if you want to find out
(x

^{3}+1/x^{3})
(x

^{3}+1/x^{3}) can be written as ((x^{2}+1/x^{2}) * (x+1/x) ) - (x+1/x)
(x

^{3}+1/x^{3}) = 14* 4 – 4 = 52
Let’s say of if you want to find out
(x

^{4}+1/x^{4})
(x

^{4}+1/x^{4}) can be written as (x^{2}+1/x^{2})^{2}-2
(x

^{4}+1/x^{4}) = 14 * 14 -2 = 194
Let’s say of if you want to find out
(x

^{5}+1/x^{5})
(x

^{5}+1/x^{5}) can be written as ((x^{3}+1/x^{3}) (x^{2}+1/x^{2})) - (x+1/x)
(x

^{5}+1/x^{5}) = 52 * 14 – 4 = 724
Let’s say of if you want to find out
(x

^{6}+1/x^{6})
(x

^{6}+1/x^{6}) can be written as (x^{3}+1/x^{3})^{2}-2
(x

^{6}+1/x^{6}) = 52 * 52 -2 = 2702
Similarly you can find out higher
power values