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# An interesting problem

ax+by=3 ; ax^2+by^2=7 ; ax^3+by^3=16 ; ax^4+bx^4=42 ; Then find ax^5+bx^5 given a,b,x,y are real nos.

Note by Avinash Panneerselvam
4 years, 11 months ago

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Solution : Note that , for any $$n >0 , (ax^n+by^n)(x+y)-xy(ax^{n-1}+by^{n-1}) = ax^{n+1}+by^{n+1}$$

So now For $$n=2$$ we get $$7(x+y)-3xy=16$$ and for $$n=3$$ we get $$16(x+y)-7xy=42$$ .
Solving gives , $$x+y=-14 , xy=-38$$

now again using the identity ,

$$ax^5+by^5 = (42)(-14)-(16)(-38) = \boxed{20}$$

- 4 years, 11 months ago

a/(ax - 1) + b/(bx - 1) = a + b . Find x .

- 2 years, 8 months ago

Is it $$ax^4 + by^4 = 42, ax^5 + by^5$$? I think you typed wrongly.

- 4 years, 11 months ago

If it is, then the answer should be $$20$$.

- 4 years, 11 months ago