Problems involving Geometric Progressions: Very Difficult Problems with Solutions

Solution:
We know that
[tex]\frac{a_{n+1}}{a_n}=2^n[/tex]
[tex]\frac{a_n}{a_{n-1}}=2^{n-1}[/tex]

[tex]\frac{a_2}{a_1}=2^{1}[/tex] We multiply all of the above in order to reach
[tex]\frac{a_{n+1}}{a_n}\cdot\frac{a_n}{a_{n-1}} \dots \frac{a_2}{a_1}=2^{n+(n-1)+\dots +1}[/tex]
[tex]\frac{a_{n+1}}{a_1}=2^{\frac{n(n+1)}{2}}[/tex], since [tex]1+2+...+n-1+n=\frac{n(n+1)}{2}[/tex]. Then
[tex]a_{n+1}=2^{\frac{n(n+1)}{2}}\cdot a_1=2^{\frac{n(n+1)}{2}}[/tex]. Substituting [tex]n+1[/tex] for [tex]n[/tex], we get
[tex]a_n=2^{\frac{n(n-1)}{2}}[/tex] and [tex]log_2(a_n)=\frac{n(n-1)}{2}[/tex]. We need [tex]log_2(a_{100})=\frac{100\cdot (100-1)}{2}=50 \cdot 99= 4950[/tex]

Solution:
[tex]a_{n+1}-a_n=3^n[/tex]
[tex]a_{n}-a_{n-1}=3^{n-1}[/tex]
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[tex]a_2-a_1=3[/tex] We add all of these in order to reach [tex]a_{n+1}-a_n+a_n-a_{n-1}+a_{n-1} \dots -a_1=3^n+3^{n-1}+\dots+3[/tex] [tex]a_{n+1}-a_1=3.(1+3+\dots + 3^{n-1})=3.\frac{3^n-1}{3-1}=\frac{3^{n+1}-3}{2}[/tex]. Substituting [tex]n+1[/tex] for [tex]n[/tex] and [tex]a_1=1[/tex], we get
[tex]a_n=\frac{3^n-3}{2}+1[/tex]. [tex]a_{10}=\frac{3^{10}-3}{2}+1=29524[/tex]

Solution:
If we add the first and second equation of the system, we get [tex]2x-y+2z+x+y+z=a+4+2a+2[/tex] or [tex]x+z=a+2[/tex]. Substituting back into the first equation, we have [tex]y+a+2=a+4[/tex], which yields [tex]y=2[/tex]. The system becomes
[tex]\begin{array}{|l}x+z=a+2\\2x+2z=2a+4\\3x+4-3z=1-2a \end{array} [/tex]. The first and second equations are linearly dependent, so we remove one of them. The system becomes
[tex]\begin{array}{|l}3x+3z=3a+6\\3x-3z=-3-2a \end{array}[/tex]. We add those two equations together and get [tex]6x=-3-2a+3a+6=a+3[/tex], or [tex]x=\frac{a+3}{6}[/tex].
By subtracting the same two equations, we get [tex]6z=3a+6+3+2a=5a+9[/tex], or [tex]z=\frac{5a+9}{6}[/tex]. Since [tex]x,2,z[/tex] form a geometric progression, we know that [tex]zx=2^2=4[/tex], or [tex]\frac{a+3}{6}.\frac{5a+9}{6}=4[/tex]
[tex](a+3)(5a+9)=144[/tex]
[tex]5a^2+24a+27=144[/tex]
[tex]5a^2+24a-117=0[/tex] : [tex]D=12^2+117\cdot 5=729=27^2[/tex], [tex]a_{1,2}=\frac{-12 \pm 27}{5}[/tex]. We know that [tex]a>0[/tex], so the negative value is discarded and the answer is [tex]a=\frac{-12+27}{5}=\frac{15}{5}=3[/tex]

Solution:
[tex](a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=364+2(ab+bc+ca)=26^2=676[/tex], so [tex]ab+bc+ca=156[/tex]. Let us denote [tex]b=ar[/tex] and [tex]c=ar^2[/tex] (geometric progressions). We have
[tex]a+b+c=a+ar+ar^2=26[/tex] and [tex]ab+bc+ca=a^2r+a^2r^3+a^2r^2=156[/tex]
[tex]a(1+r+r^2)=26[/tex] and [tex]a^2r(1+r+r^2)=156=6\cdot 26[/tex]. We divide both equations and get
[tex]\frac{a^2r(1+r+r^2)}{a(1+r+r^2)}=6[/tex], or [tex]ar=b=6[/tex].

Solution:
We shall generalize the problem: find the infinite sum [tex]S=1+2x+3x^2+4x^3+...[/tex] for [tex]|x| < 1[/tex].
[tex]1+2x+3x^2+4x^3+...=(1+x+x^2+x^3+...)+x+2x^2+3x^3+...=\\=(1+x+x^2+x^3+...)+(x+x^2+x^3+...)+x^2+2x^3+3x^4+...=1.(1+x+x^2+x^3+...)+x(1+x+x^2+x^3+...)+x^2(1+x+x^2+x^3+...)+...=\\=(1+x+x^2+x^3+...)(1+x+x^2+x^3+...)=(1+x+x^2+x^3+...)^2[/tex]. We can take out x behind the parentheses only because there are infinitely many terms.
But [tex]1+x+x^2+x^3+...[/tex] is the sum of an infinite geometric series with [tex]a_1=1[/tex] and [tex]r=x[/tex] and its sum is [tex]1+x+x^2+x^3+...=1.\frac{1}{1-x}=\frac{1}{1-x}[/tex], therefore [tex]S=(\frac{1}{1-x})^2[/tex]. Plugging back [tex]x=\frac{1}{7}[/tex], we get [tex]S=(\frac{1}{1-\frac{1}{7}})^2=(\frac{1}{\frac{6}{7}})^2=(\frac{7}{6})^2=\frac{49}{36}[/tex]