Question

f(x + 3) = f(x + 1) + 4x + 12 and f(0) = 2, then what is f(300)?​

Answer 1

Composite function $g(x)=x-3$

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$f(x)=f(x-2)+4(x-3)+12$

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$f(x)-f(x-2)=4x-12+12$

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$f(x)-f(x-2)=4x$

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Composite function $h(x)=2x$

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$f(2x)-f(2x-2)=8x$

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Now, recall the properties of the telescoping series. The sequences alternate, thus leaving the difference of the first and last term.

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The property of,

$\cdots\longrightarrow\boxed{\displaystyle\sum^{n}_{k=1}(a_{k+1}-a_{k})=a_{n+1}-a_{1}}$

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$\displaystyle\sum^{n}_{x=1}\{f(2x)-f(2x-2)\}=\sum^{n}_{x=1}8x$

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$\displaystyle\sum^{n}_{x=1}\{f(2x)-f(2x-2)\}=8\times\dfrac{n(n+1)}{2}$

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$\displaystyle\sum^{n}_{x=1}\{f(2x)-f(2x-2)\}=4n(n+1)$

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$\displaystyle\sum^{n}_{x=1}\{f(2x)-f(2x-2)\}=4n^{2}+4n$

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$2n=300$

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$n=150$

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$\small\displaystyle\sum^{150}_{x=1}\{f(2x)-f(2x-2)\}=4\cdot 150^{2}+4\cdot 150$

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$f(300)-f(0)=90000+600$

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$f(300)=90600+2$

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$\therefore f(300)=90602$