Question 199771
You basically have the sequence: 
12*100,
12*200,
12*300,
12*400,
12*500, ...



note: I'm multiplying by 12 since he deposits money 12 times a year.



and you're adding up the terms to get: 
12*100+12*200=12*300,
12*100+12*200+12*300=12*600,
12*100+12*200+12*300+12*400=12*1000,
12*100+12*200+12*300+12*400+12*500=12*1500, ...




and you want to know which partial sum will be equal to 66,000. 



So you basically want to know when 12*100+12*200+12*300+12*400+12*500+...=66000



We can factor out the GCF 12*100 to get: 12*100(1+2+3+4+5+....)=66000



and we're left with a very familiar series in the parenthesis.



It turns out that *[Tex \LARGE \displaystyle 1+2+3+4+5+\ldots+n=\sum_{k=1}^n k = \frac{n(n+1)}{2}] (this is a well known formula).



So the equation then becomes: {{{12*100((n(n+1))/2)=66000}}}



where "n" is the number of years.



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{{{12*100((n(n+1))/2)=66000}}} Start with the given equation.



{{{100((n(n+1))/2)=66000/12}}} Divide both sides by 12.



{{{100((n(n+1))/2)=5500}}} Reduce



{{{100(n(n+1))/2=5500/100}}} Divide both sides by 100.



{{{(n(n+1))/2=55}}} Reduce.



{{{n(n+1)=55*2}}} Multiply both sides by 2.



{{{n(n+1)=110}}} Multiply



{{{n^2+n=110}}} Distribute



{{{n^2+n-110=0}}} Subtract 110 from both sides.



Notice we have a quadratic in the form of {{{An^2+Bn+C}}} where {{{A=1}}}, {{{B=1}}}, and {{{C=-110}}}



Let's use the quadratic formula to solve for "n":



{{{n = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{n = (-(1) +- sqrt( (1)^2-4(1)(-110) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=1}}}, and {{{C=-110}}}



{{{n = (-1 +- sqrt( 1-4(1)(-110) ))/(2(1))}}} Square {{{1}}} to get {{{1}}}. 



{{{n = (-1 +- sqrt( 1--440 ))/(2(1))}}} Multiply {{{4(1)(-110)}}} to get {{{-440}}}



{{{n = (-1 +- sqrt( 1+440 ))/(2(1))}}} Rewrite {{{sqrt(1--440)}}} as {{{sqrt(1+440)}}}



{{{n = (-1 +- sqrt( 441 ))/(2(1))}}} Add {{{1}}} to {{{440}}} to get {{{441}}}



{{{n = (-1 +- sqrt( 441 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{n = (-1 +- 21)/(2)}}} Take the square root of {{{441}}} to get {{{21}}}. 



{{{n = (-1 + 21)/(2)}}} or {{{n = (-1 - 21)/(2)}}} Break up the expression. 



{{{n = (20)/(2)}}} or {{{n =  (-22)/(2)}}} Combine like terms. 



{{{n = 10}}} or {{{n = -11}}} Simplify. 



So the <i>possible</i> solutions are {{{n = 10}}} or {{{n = -11}}} 

  

Recall that the value of "n" is the number of years. Since you cannot have a negative year, this excludes {{{n = -11}}} 



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Answer:



So the answer is {{{n = 10}}} which means that it will take 10 years to save up to $66,000.



If you have any questions, email me at jim_thompson5910@hotmail.com