# Problem 1.

In-1 Find $f(1)+f(2)+f(3)+\ldots+f(13)$, if $f(n)=4 n^{3}-6 n^{2}+4 n+13$.
Answer: 28743.

Solution. Since $f(n)=n^{4}-(n-1)^{4}+14$, then

$f(1)+f(2)+f(3)+\ldots+f(13)=1^{4}+2^{4}+\ldots+13^{4}-\left(0^{4}+1^{4}+\ldots+12^{4}\right)+14 \cdot 13=13^{4}+14 \cdot 13=28743$.

V-2 Find $f(1)+f(2)+f(3)+\ldots+f(11)$, if $f(n)=4 n^{3}+6 n^{2}+4 n+9$.

Answer: 20823.

V-3 Find $f(1)+f(2)+f(3)+\ldots+f(12)$, if $f(n)=4 n^{3}-6 n^{2}+4 n+7$.

Answer: 20832.

V-4 Find $f(1)+f(2)+f(3)+\ldots+f(14)$, if $f(n)=4 n^{3}+6 n^{2}+4 n+5$.

Answer: 50680.

V-5 Let $f(x)=x^{2}+10 x+20$. Solve the equation

$$
f(f(f(f(f(x)))))=0
$$

Answer: $-5 \pm \sqrt[32]{5}$.

Solution. Note that $f(x)=(x+5)^{2}-5$. Therefore, $f(f(x))=(x+5)^{4}-5, f(f(f(x)))=(x+5)^{8}-5, \ldots, f(f(f(f(f(x)))))=(x+5)^{32}-5=0$, from which $x=-5 \pm \sqrt[32]{5}$.

V-6 Let $f(x)=x^{2}+6 x+6$. Solve the equation

$$
f(f(f(f(f(x)))))=0
$$

Answer: $-3 \pm \sqrt[32]{3}$.

V-7 Let $f(x)=x^{2}+14 x+42$. Solve the equation

$$
f(f(f(f(x))))=0
$$

Answer: $-7 \pm \sqrt[16]{7}$.

V-8 Let $f(x)=x^{2}+18 x+72$. Solve the equation

$$
f(f(f(f(x))))=0
$$

Answer: $-9 \pm \sqrt[16]{9}=-9 \pm \sqrt[8]{3}$.

## Lomonosov Olympiad for Schoolchildren in Mathematics

Final Stage 2020/21 Academic Year for 10-11 Grades

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