Home Pre-Algebra Simplify expressions of real numbers and fractions

Simplify expressions of real numbers and fractions

We teach you how to simplify the expressions of real numbers. In fact, mathematical calculation of fractions and complicated math, formulas are often considered difficult tasks for students. One of the best exercises for middle and high school students is to simplify real number expressions and solve inequalities. This requires knowledge of the categories of numbers and their properties. Here, this article presents an overview of the rules and properties of numbers, then applies them to calculate complicated expressions and numbers.

Exercises on how to simplify expressions of real numbers

Exercise: Show that for any $n\in\mathbb{N}$ \begin{align*}\tag{$P_n$} (n+1)!\ge \sum_{k=1}^n k!. \end{align*}

Solution: For $n=1,$ we have $(1+1)!=2!ge 1!,$ hence the property $(P_1)$ is vérified. Assume now, by induction, that $(P_n)$ holds. As $n+2>2,$ then \begin{align*}\tag{1} (n+2)!=(n+2)(n+1)!\ge 2(n+1)!.\end{align*}
On the other hand, by adding $(n+1)!$ to the both sides of the inequality $(P_n),$ we obtain \begin{align*}\tag{2}2(n+1)!\ge \sum_{k=1}^nk!+(n+1)!=\sum_{k=1}^{n+1}k!\end{align*}
By combining (1) et (2), we obtain \begin{align*} (n+2)!\ge \sum_{k=1}^{n+1}k!.\end{align*}
Thus $(P_{n+1})$ holds.

Exercise: Let $x\in \mathbb{R}$ and $n\in\mathbb{N}^\ast$. Calculate the sum \begin{align*} S_n(x)= \sum_{k=0}^n x^k. \end{align*} Deduce the value of \begin{align*} R_n(x)=\sum_{k=0}^n k x^k. \end{align*}

Proof: For any $x\in \mathbb{R},$ we have $(x-1)S_n(x)=\sum_{k-0}^n x^{k+1}-S_n(x)$. By making the change of index $m=k+1,$ we get   \begin{align*}\sum_{k=1}^{n+1} k x^{k+1}-S_n(x)= (-1+S_n(x)+x^{n+1})-S_n(x)=x^{n+1}-1.\end{align*} Hence, for any $x\in\mathbb{R}\setminus\{1\},$ we have $$S_n(x)=\frac{x^{n+1}-1}{x-1}.$$ On the other hand, the function $x\in \mathbb{R}\setminus\{1\}\mapsto S_n(x)$ is differentiabl, and \begin{align*} \frac{d}{dx}S_n(x)=\frac{d}{dx}\frac{x^{n+1}-1}{x-1}.\end{align*} As $S_n(x)$ is defined by a finie sum, then we have \begin{align*} \frac{d}{dx}S_n(x)&=\sum_{k=0}^n k \frac{d}{dx}(x^k)= \sum_{k=1}^n k x^{k-1}=\sum_{k=0}^{n-1} (k+1)x^k\cr &= S_{n-1}(x)-nx^n+R_n(x).\end{align*} Thus $$R_n(x)=\frac{d}{dx}S_n(x)-\left( S_{n-1}(x)-nx^n\right).$$ Now we compute \begin{align*} S_{n-1}(x)-nx^n&=\frac{x^n-1}{x-1}-nx^{n}\cr &= \frac{(n+1)x^n-nx^{n+1}-1}{x-1}.\end{align*} On the other hand, \begin{align*} \frac{d}{dx}S_n(x)&=\left( \frac{x^{n+1}-1}{x-1}\right)’\cr &= \frac{(n+1)x^n (x-1)-(x^{n+1}-1)}{(x-1)^2}.\end{align*} An elementary calculus shows that \begin{align*}R_n(x)=\frac{-nx^{n+2}-(n+1)x^{n+1}-x}{(x-1)^2}.\end{align*}

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