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horsena [70]
2 years ago
14

Given that the following two geometric series(meetkundige reekse) are covergent 1+x+x²+x³+... And 1-x+x²-x³+... Determine the va

lue(s) of x for witch the sum of the following two series is equel​
Mathematics
1 answer:
Oksanka [162]2 years ago
3 0

The sets are (1,-1) or (-1,1)

Lets see

Take until 4terms

#1

\\ \rm\longmapsto S_4=1+1+1^2+1^3=1+1+1+1=4(1,-1)

\\ \rm\longmapsto S_4=1+(-1)+(-1)^2+(-1)^3=1-1+1-1=0(-1,1)

#2

\\ \rm\longmapsto S_4=1-(-1)+(-1)^2-(-1)^3=1+1+1+1=4(1,-1)

\\ \rm\longmapsto S_4=1-1+(1)^2-1^3=1-1+1-1=0(-1,1)

Hence verified

Another set can be (0,0)

Proof:-

\\ \ast\rm\longmapsto S_3=1+0+0^2=1

\\ \ast\rm\longmapsto S_3=1-0+0^2=1

Hence verified.

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If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years f
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Part A) Annual \$66,480.95  

Part B) Semiannual \$66,862.38  

Part C) Monthly \$67,195.44  

Part D) Daily \$67,261.54  

Step-by-step explanation:

we know that

The compound interest formula is equal to  

A=P(1+\frac{r}{n})^{nt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

so

Part A) Annual

in this problem we have  

t=5\ years\\ P=\$47,400\\ r=0.07\\n=1  

substitute in the formula above  

A=\$47,400(1+\frac{0.07}{1})^{1*5}  

A=\$47,400(1.07)^{5}  

A=\$66,480.95  

Part B) Semiannual

in this problem we have  

t=5\ years\\ P=\$47,400\\ r=0.07\\n=2  

substitute in the formula above  

A=\$47,400(1+\frac{0.07}{2})^{2*5}  

A=\$47,400(1.035)^{10}  

A=\$66,862.38  

Part C) Monthly

in this problem we have  

t=5\ years\\ P=\$47,400\\ r=0.07\\n=12  

substitute in the formula above  

A=\$47,400(1+\frac{0.07}{12})^{12*5}  

A=\$47,400(1.0058)^{60}  

A=\$67,195.44  

Part D) Daily

in this problem we have  

t=5\ years\\ P=\$47,400\\ r=0.07\\n=365  

substitute in the formula above  

A=\$47,400(1+\frac{0.07}{365})^{365*5}  

A=\$47,400(1.0002)^{1,825}  

A=\$67,261.54  

8 0
3 years ago
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