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Doubt 4


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Note by Radhesh Sarma
1 year, 10 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. Krishna Sharma · 1 year, 10 months ago

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@Krishna Sharma @Krishna Sharma ,thanks a lot Radhesh Sarma · 1 year, 10 months ago

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