How to work: choose the level you start from. If you solve the problems of a level without trouble, move on to the next one. First reread the theory in the worksheet: the properties (P1)–(P3) and the standard sums (S1)–(S4).
Remember
$\sum_{k=1}^{n}{c}=n\cdot c$ $\sum_{k=1}^{n}{k}=\frac{n(n+1)}{2}$ $\sum_{k=1}^{n}{k^2}=\frac{n(n+1)(2n+1)}{6}$ $\sum_{k=1}^{n}{k^3}=\left(\frac{n(n+1)}{2}\right)^2$
Induction: Step I $P(n_0)$ is true; Step II $P(k)\Rightarrow P(k+1)$; conclusion: $P(n)$ is true for every $n\ge n_0$.
Basic level
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Write out, then calculate:
a) $\sum_{k=1}^{6}{(3k-1)}$;
b) $\sum_{k=0}^{4}{2^k}$;
c) $\sum_{k=3}^{7}{(k^2-1)}$;
d) $\sum_{k=1}^{100}{\left(-1\right)^k}$.
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Write with the Σ symbol, after you find the general term:
a) $5+10+15+\ldots+200$;
b) $1+8+27+\ldots+1000$;
c) $2+6+10+\ldots+(4n-2)$;
d) $1\cdot 3+2\cdot 5+3\cdot 7+\ldots+n(2n+1)$;
e) $\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots+\frac{1}{3^n}$.
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Calculate, using the standard sums:
a) $1+2+\ldots+200$;
b) $1^2+2^2+\ldots+20^2$;
c) $1^3+2^3+\ldots+10^3$;
d) $2+4+6+\ldots+100$.
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Calculate, using the properties of sums:
a) $\sum_{k=1}^{n}{(6k+1)}$;
b) $\sum_{k=1}^{n}{(2k-3)}$;
c) $\sum_{k=1}^{10}{(k^2+2k)}$.
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Prove by mathematical induction that, for every $n\in \N^*$:
a) $2+4+6+\ldots+2n=n(n+1)$;
b) $1+4+7+\ldots+(3n-2)=\frac{n(3n-1)}{2}$.
Consolidated level
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Write with Σ, then calculate:
a) $3+5+7+\ldots+(2n+1)$;
b) $2^2+4^2+6^2+\ldots+(2n)^2$;
c) $1\cdot 5+2\cdot 6+3\cdot 7+\ldots+n(n+4)$.
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Let $S_n=1\cdot 2+3\cdot 4+5\cdot 6+\ldots+(2n-1)\cdot 2n$.
a) Write $S_n$ with Σ and calculate it.
b) Prove by induction the formula you found in a).
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Calculate:
a) $\sum_{k=11}^{30}{k}$;
b) $\sum_{k=5}^{10}{k^3}$;
c) $\sum_{k=n+1}^{2n}{k^2}$.
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Find the nonzero natural number $n$ for which:
a) $1+2+\ldots+n=78$;
b) $1+3+5+\ldots+(2n-1)=169$;
c) $1^2+2^2+\ldots+n^2=204$.
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Prove by mathematical induction that, for every $n\in \N^*$:
a) $1^2+3^2+5^2+\ldots+(2n-1)^2=\frac{n(2n-1)(2n+1)}{3}$;
b) $1+2+2^2+\ldots+2^{n-1}=2^n-1$.
Advanced level
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A telescoping sum.
a) Check the equality $\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}\cdot \left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)$.
b) Calculate $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\ldots+\frac{1}{(2n-1)(2n+1)}$.
c) Prove the result of b) by induction.
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Calculate $\sum_{k=1}^{n}{k(n-k)}$, for $n\ge 2$. Watch out: in this sum, $n$ does not depend on $k$. Check the result for $n=3$.
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Prove by induction that $\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\ldots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$, for every $n\in \N^*$.
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Find the mistake. A student calculates like this: “$\sum_{k=1}^{n}{k(k+1)}=\sum_{k=1}^{n}{k}\cdot \sum_{k=1}^{n}{(k+1)}$”.
a) Check the equality for $n=2$.
b) Which property of sums did the student apply wrongly?
c) Calculate the sum correctly.
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Challenge. Calculate $1^2-2^2+3^2-4^2+5^2-6^2$ and $1^2-2^2+3^2-4^2+5^2-6^2+7^2$. What do you notice? Prove by induction that, for every $n\in \N^*$:
$$\sum_{k=1}^{n}{\left(-1\right)^{k+1}\cdot k^2}=\left(-1\right)^{n+1}\cdot \frac{n(n+1)}{2}$$