Question

(x+1/x)^2 + (x^2 +1/x^2)^2+………(x^27 + 1/x^27)^2 = ?

Answer 1

If G is the geometric mean of X & Y, then (1/(g^2-x^2)) + (1/(g^2-y^2)) = ?

Answer 2

Hey!!! plz help me.... trying since so many days n recently bumped into this web site... plz let me know the solution

Answer 3

Your first question:

(1) \[\large{(x+\cfrac{1}{x})^2 + (x^2 + \cfrac{1}{x^2})^2 + ...+(x^{27} + \cfrac{1}{x^{27}})^2}\]

\[\large{= \sum_{i = 1}^{27}(x^i + \cfrac{1}{x^i})^2}\]

\[\large{= \sum_{i=1}^{27}((x^i)^2 + 2\times x^i\times \cfrac{1}{x^i} + (\cfrac{1}{x^i})^2)}\]

\[\large{= \sum_{i=1}^{27}(x^{2i}) + \sum_{i=1}^{27}2 + \sum_{i=1}^{27}(\cfrac{1}{x^{2i}})}\]

\[\large{= S_1 + 2\times 27 + S_2 = S_1 + S_2 + 54}\]

Where,

\[\large{S_1 = \sum_{i=1}^{27} (x^{2i})}\]

\[\large{S_2 = \sum_{i=1}^{27}(\cfrac{1}{x^{2i}})}\]

Do you have any doubt till this step ?

Answer 4

i mean its clear.... thnx a bunch

Answer 5

If G is the geometric mean of X and Y, then 1/(G^2 – X^2) + 1/(G^2 – Y^2) =?
How to answer this question?.... not getting.

Answer 6

@Jemson ,

Geometric mean of numbers X and Y ,

\[\large{G = \sqrt{XY}}\]

Thus,

\[\large{\cfrac{1}{G^2 - X^2} + \cfrac{1}{G^2 - Y^2}}\]
\[\large{= \cfrac{G^2 - Y^2 + G^2 - X^2}{(G^2-X^2)(G^2-Y^2)}}\]
\[\large{= \cfrac{2G^2-(X^2+Y^2)}{(G^2-X^2)(G^2-Y^2)}}\]
\[\large{= \cfrac{2(XY) - (X^2+Y^2)}{(XY-X^2)(XY-Y^2)}}\]
\[\large{= \cfrac{-(X^2+Y^2-2XY)}{X(Y-X)Y(X-Y)}}\]
\[\large{= \cfrac{-(X-Y)^2}{-XY(X-Y)^2}}\]
\[\large{= \cfrac{1}{XY}}\]

Answer 7

Tell me if you can't understand any step. Okay ?