If x+1x=3 then find x5+1x5.
Solve for x5+1x5
It is given that x+1x=3
Squaring on both sides
⇒x+1x2=9⇒x2+1x2+2=9
⇒x2+1x2=7…………..(1)
Similarly
⇒x+1x3=27⇒x3+1x3+3x21x+3x1x2=27⇒x3+1x3+3x+31x=27⇒x3+1x3+3x+1x=27⇒x3+1x3+3×3=27⇒x3+1x3=27-9
x3+1x3=18…………(2)
Multiply equation (1) and (2)
⇒x2+1x2×x3+1x3=18×7⇒x5+1x5+x+1x=126⇒x5+1x5=126-3⇒x5+1x5=123
Hence the value of x5+1x5is 123
If f=x1+x2+13(x1+x2)3+15(x1+x2)5+... to ∞ and g=x−23x3+15x5+17x7−29x9+..., then f=d×g. Find 4d.