If `x^2-x+1=0` then the value of `sum_[n=1]^[5][x^n+1/x^n]^2` is:
127 views · Published 11 October 2018 · 6:53 · Indexed 29 September 2026
Channel: Doubtnut · 2018 · Education
To ask Unlimited Maths doubts download Doubtnut from - https://goo.gl/9WZjCW If `x^2-x+1=0` then the value of `sum_[n=1]^[5][x^n+1/x^n]^2` is:
More from this channel
-
4:30
IIT JEE APPLICATION OF DERIVATIVES Find the equation of tangent to the curve
`y=sin^(-1)(2x)/(1+...
-
2:28
IIT JEE LIMITS AND DERIVATIVES Evaluate:
`("lim")_(xrarr0)(sinx^0)/x`
-
4:29
IIT JEE LIMITS AND DERIVATIVES Evaluate the limit:
`("lim")_(xvecoo)((x+2)/(x+1))^(x+3)`
-
3:50
`A D` and `B E` are respectively altitudes of an isosceles triangle `A B C` with `A C=B C` . Pro...
-
4:19
IIT JEE PERMUTATIONS AND COMBINATIONS Find the number of distinct rational numbers `x`
such that ...
-
2:15
IIT JEE PERMUTATIONS AND COMBINATIONS Find the sum of the series `sum_(r=1)^n rxxr !dot`
-
2:47
JEE MAINS 2018 Find the points of local maxima and local minima,
if any, of each of the follo...
-
3:03
Find the number of polynomials of the form `x^3+a x^2+b x+c`
that are divisible by `x^2+1,w h e...