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# Doubt 4

The infinite geometric series $$\displaystyle\sum_{r=0}^\infty \frac9{(y^2-4y+5)^{r-1}}$$ where $$y=x^2-6x+11$$ has a finite sum then $$x$$ cannot be equal to:

1 year, 3 months ago

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Rewrite $$(y^2 - 4y + 5)$$ as $$\displaystyle( (y-2)^2 + 1)$$

Now substitute the value of $$y$$ in terms of $$x$$ to get(I am just writing the denominator part)

$$\displaystyle ((x^2 - 6x + 9)^2 + 1)^{r-1}$$

Observe that the quadratic of $$x$$ is a perfect square of $$(x-3)$$ hence the expression reduces to

$$\displaystyle ((x-3)^4 + 1)^{r-1}$$

Since the sum should converge the common ratio of the G.P should lie between -1 and 1 And clearly sum will not converge when $$x=3$$ because we will get $$9+9+9+9 + \ldots$$ infinite times. · 1 year, 3 months ago