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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?

2 Answers 2

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$f(2)=3$ by the substitution $x=y=2$. Then $f(1)=1$, by the substitution $x=y=1$. Then, $f(n+1)-f(n)=f(1)+1\cdot n=1+n$.

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    may u plz explain what u did i did not get it @leo163 – 2017-01-24
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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 $f(n+1)-f(n)=1+n$.

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    may u plz explain what u did i did not get it @myglasses – 2017-01-24
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    first set in formula $x=y=2$ and next $x=y=1$ – 2017-01-24
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    what do u mean? – 2017-01-24
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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 – 2017-01-24
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    for factor $xy$ – 2017-01-24
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    $f(x + y) = f(x) + f(y) + xy$ so $f(2 + 2) = f(2) + f(2) + 2\times2$. – 2017-01-24
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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 – 2017-01-24
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    This is mysterious. – 2017-01-24
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    ok thanks... i figured it out by myself... u dont actually need f(4)=10 – 2017-01-24
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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..!!!:):):) – 2017-01-24