IIT JEE , AIEEE Forum - Mathematics , Algebra

nadeemoidu

@ryuamakusa

I'll quote the lines where u went wrong

>>y₁²-2y₁+1 + y₂²-2y₂+4 +....... y_n² -2y_n + n²=0

>>now you can group the terms as

>>(y₁-1)² + (y₂-2)² +.......(y_n-n)²=0 ------------(1)

Do think that this step is correct. Can u get that expression by opening the brackets?

RyuAmakusa

well I am sorry! nadeemoidu i just over looked an very glaring error! thank's for correcting.but you see i continued the same way and got there was no. real sol.

is something wrong.

hsbhatt

Actually rohit, your method is right and very elegant. You stumbled only because you've written the substitution wrongly. It should be $x_i = i^2 \sec^2{\theta_i}$

Then, when you conclude that $\tan{\theta_i} = 1$ , then you get $\sec^2{\theta_i} = 2$ and hence x_i = 2i^2

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