A certain function $f$ satisfies $f(x + y) = f(x) + f(y) + xy$ for all real numbers $x$ and $y$, and it is known that $f(4) = 10$. Then $f(n+1) − f(n)$ is equal to:
How do we solve this?
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$f(2)=3$ by the substitution $x=y=2$. Then $f(1)=1$, by the substitution $x=y=1$. Then, $f(n+1)-f(n)=f(1)+1\cdot n=1+n$.
Leo163
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may u plz explain what u did i did not get it @leo163 – user408113 Jan 24 '17 at 20:36
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$f(4)=f(2)+f(2)+4=(f(1)+f(1)+1)+(f(1)+f(1)+1)+4=4f(1)+6=10$ so $f(1)=1$ and $f(n+1)=f(n)+f(1)+n$ thus $f(n+1)-f(n)=1+n$.
Nosrati
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can u plz be clearer with ur notation?? i get that f(4)=f(2)+f(2) but why +4, where did u get that from – user408113 Jan 24 '17 at 20:42
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oh ok i get it thank u...... but why did u choose 2 and 1??? whats sig about these 2 numbers – user408113 Jan 24 '17 at 20:58
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ok thanks... i figured it out by myself... u dont actually need f(4)=10 – user408113 Jan 24 '17 at 21:03
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if u expand f(n+1)-f(n) in the beginig u get: f(n)+1+n-f(n), so f(n) cancel and u just get n+1..!!!:):):) – user408113 Jan 24 '17 at 21:04