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Home » This is a telescoping series. Writing the partial sum \( S_N \): - AMAZONAWS

This is a telescoping series. Writing the partial sum \( S_N \): - AMAZONAWS

📅 March 5, 2026 👤 scraface

Mar 05, 2026

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📌 Substitute \( (1, 0) \): \( 0 = a(1-2)^2 - 3 \).

📌 Simplifying, \( 0 = a(1) - 3 \), so \( a = 3 \).

📌 Expand to standard form: \( f(x) = 3(x-2)^2 - 3 = 3(x^2 - 4x + 4) - 3 = 3x^2 - 12x + 12 - 3 \).

📌 Thus, \( f(x) = 3x^2 - 12x + 9 \), so \( a = 3 \), \( b = -12 \), \( c = 9 \).

📌 #### \( a = 3, b = -12, c = 9 \)

📌 The sum of an infinite geometric series is 5, and the first term is 2. Find the common ratio.

📌 The sum \( S \) of an infinite geometric series is \( S = rac{a}{1-r} \).

📌 Given \( S = 5 \) and \( a = 2 \), we have \( rac{2}{1-r} = 5 \).

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We aim to find $ f(x^2 - 2) $.
First, observe that the right-hand side can be expressed in terms of $...
x^4 + 4x^2 + 3 = (x^2)^2 + 4x^2 + 3.
Let $ u = x^2 + 2 $. Then $ x^2 = u - 2 $, and
Substituting into the expression:
x^4 + 4x^2 + 3 = (u^2 - 4u + 4) + 4(u - 2) + 3 = u^2 - 4u + 4 + 4u - 8...
f(u) = u^2 - 1.
Now substitute $ u = x^2 - 2 $ into $ f $:
f(x^2 - 2) = (x^2 - 2)^2 - 1 = x^4 - 4x^2 + 4 - 1 = x^4 - 4x^2 + 3.
\boxed{x^4 - 4x^2 + 3}

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