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Sagot :
[tex]a_n=5 \ for \ n=1 \\
a_n=-10 \ for \ n=2 \\
a_n=20 \ for \ n=3[/tex]
You can see it's a geometric sequence because the ratio is constant ([tex]\frac{a_3}{a_2}=\frac{a_2}{a_1}[/tex]).
The nth term of a geometric sequence:
[tex]a_n=a_1 \times r^{n-1}[/tex]
[tex]a_1=5 \\ r=\frac{a_2}{a_1}=\frac{-10}{5}=-2 \\ \\ a_n=5(-2)^{n-1}[/tex]
It has to be the sum for term 4 through term 15, so n=4 and the number above the sigma is 15.
Your answer:
[tex]\sum^{15}_{n=4} 5(-2)^{n-1}[/tex]
You can see it's a geometric sequence because the ratio is constant ([tex]\frac{a_3}{a_2}=\frac{a_2}{a_1}[/tex]).
The nth term of a geometric sequence:
[tex]a_n=a_1 \times r^{n-1}[/tex]
[tex]a_1=5 \\ r=\frac{a_2}{a_1}=\frac{-10}{5}=-2 \\ \\ a_n=5(-2)^{n-1}[/tex]
It has to be the sum for term 4 through term 15, so n=4 and the number above the sigma is 15.
Your answer:
[tex]\sum^{15}_{n=4} 5(-2)^{n-1}[/tex]
Answer:
the summation of 4 times negative 3 to the n minus 1 power, from n equals 4 to 15.
Step-by-step explanation:
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