Sequence AND Series flash cards

1. 1.For an AP with first term $a$ and common difference $d$, give the formulas for the $n^{th}$ term $T_n$ and the sum of first $n$ terms $S_n$.

   $T_n = a+(n-1)d$ and $S_n = \dfrac{n}{2}[2a+(n-1)d] = \dfrac{n}{2}(a+T_n)$

   Hint: Think linear growth vs. average of first and last term.

2. 2.For a GP with first term $a$ and common ratio $r$, what is $S_n$ (for $r\neq1$) and $S_\infty$ (for $|r|<1$)?

   $S_n = \dfrac{a(r^n-1)}{r-1}$ and $S_\infty = \dfrac{a}{1-r}$

   Hint: Infinite sum needs the ratio's absolute value below 1.

3. 3.State the relation between Arithmetic Mean $(A)$, Geometric Mean $(G)$ and Harmonic Mean $(H)$ of two positive numbers, and the AM-GM-HM inequality.

   $G^2=AH$ and $A\geq G\geq H$, with equality iff all numbers are equal.

   Hint: G is the geometric mean between A and H.

4. 4.Give the standard summation formulas for $\sum n$, $\sum n^2$, and $\sum n^3$ (first $n$ natural numbers).

   $\sum n=\dfrac{n(n+1)}{2}$, $\sum n^2=\dfrac{n(n+1)(2n+1)}{6}$, $\sum n^3=\left[\dfrac{n(n+1)}{2}\right]^2$

   Hint: Note $\sum n^3 = (\sum n)^2$.

5. 5.What method is used to sum an Arithmetico-Geometric Progression (A.G.P.) like $a, (a+d)r, (a+2d)r^2,\dots$, and what is $S_\infty$ for $|r|<1$?

   Multiply $S_n$ by $r$ and subtract term-by-term; $S_\infty=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}$

   Hint: Called the $S - rS$ technique.

6. 6.If $a,b,c>0$ and $a+b+c=1$, find the minimum value of $\dfrac1a+\dfrac1b+\dfrac1c$.

   By the AM-HM inequality, $\dfrac{a+b+c}{3}\ge\dfrac{3}{\frac1a+\frac1b+\frac1c}$, which rearranges to $\dfrac1a+\dfrac1b+\dfrac1c\ge\dfrac{9}{a+b+c}=9$. Equality holds when $a=b=c=\dfrac13$. So the minimum value is $9$.

   Hint: Apply the AM-HM inequality directly to $a,b,c$.

7. 7.For $a,b,c>0$, prove that $(a+b)(b+c)(c+a)\ge 8abc$, and state when equality holds.

   By AM-GM, $a+b\ge2\sqrt{ab}$, $b+c\ge2\sqrt{bc}$, $c+a\ge2\sqrt{ca}$. Multiplying these three inequalities gives $(a+b)(b+c)(c+a)\ge 8\sqrt{a^2b^2c^2}=8abc$. Equality holds iff $a=b=c$.

   Hint: Apply AM-GM separately to each of the three factors, then multiply.

8. 8.Find the minimum value of $(a+\frac1b)(b+\frac1c)(c+\frac1a)$ for $a,b,c>0$.

   By AM-GM, $a+\dfrac1b\ge2\sqrt{a/b}$, and similarly for the other two factors. Multiplying: $(a+\frac1b)(b+\frac1c)(c+\frac1a)\ge 8\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{a}}=8$. Equality holds at $a=b=c=1$, so the minimum value is $8$.

   Hint: Bound each of the three factors below using AM-GM, then multiply.

9. 9.Prove that for $a,b,c>0$, $a^3+b^3+c^3\ge 3abc$, with equality iff $a=b=c$.

   Using $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, note the second factor equals $\tfrac12\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\ge0$. Since $a+b+c>0$, the whole product is $\ge0$, giving $a^3+b^3+c^3\ge3abc$. Equality forces all three squares to vanish, i.e. $a=b=c$.

   Hint: Factor $a^3+b^3+c^3-3abc$ and examine the sign of each factor.

10. 10.Find the minimum value of $\dfrac{x^2+3x+3}{x+1}$ for $x>-1$.

    Substitute $t=x+1>0$. Then $x^2+3x+3=(t-1)^2+3(t-1)+3=t^2+t+1$, so the expression becomes $\dfrac{t^2+t+1}{t}=t+1+\dfrac1t$. By AM-GM, $t+\dfrac1t\ge2$, so the expression is $\ge2+1=3$. Equality at $t=1$, i.e. $x=0$. Minimum value $=3$.

    Hint: Substitute $t=x+1$ to reduce the expression to $t+\dfrac1t$ plus a constant.

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