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Ivahew [28]
4 years ago
14

Evaluate the series 4+2+1+1/2+1/4 to s10

Mathematics
2 answers:
stiks02 [169]4 years ago
8 0

Answer:

<u>The correct answer is 13.75</u>

Step-by-step explanation:

4+2= 6

6+1=7

6+7= 13

13+1/2= 13.5

13.5+1/4= 13.75

Alja [10]4 years ago
3 0

Answer:

s_10=\frac{1023}{128}

Step-by-step explanation:

4+2+1+1/2+1/4.....

WE need to find out the sum s10

first term is 4 and second term is 2

2/4 = 1/2

1/2= 1/2

1/2 / 1= 1/2

so common ratio is 1/2

The given series is geometric series

we use the sum formula for the geometric series

s_n= a\frac{(1-r^n)}{1-r}

Where 'r' is the common ratio and 'a' is the first term

a= 4  and r= 1/2

Plug in the values in the sum formula'

s_n= 4\frac{(1-(\frac{1}{2})^n)}{1-(\frac{1}{2})}

Given n=10

s_10= 4\frac{(1-(\frac{1}{2})^{10})}{1-(\frac{1}{2})}=\frac{1023}{128}

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Find the first five terms of the sequence in which a1 = 6 and an = –3an – 1 – 12, if n ≥ 2.
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Answer:

\huge\boxed{6,\ -30;\ 78;\ -246;\ 690}

Step-by-step explanation:

a_1=6\\a_n=-3a_{n-1}-12\qquad n\geq2\\\\\text{Substitute}\ n=2,\ n=3,\ n=4,\ n=5\ \text{to}\ a_n=-3a_{n-1}-12:\\\\\text{for}\ n=2\\\\a_2=-3a_{2-1}-12=-3a_1-12\\\\\text{substitute}\ a_1=6\\\\a_2=-3(6)-12=-18-12=-30\\\\\text{for}\ n=3\\\\a_3=-3a_{3-1}-12=-3a_2-12\\\\\text{substitute}\ a_2=-30\\\\a_3=-3(-30)-12=90-12=78

\text{for}\ n=4\\\\a_4=-3a_{4-1}-12=-3a_3-12\\\\\text{substitute}\ a_3=78\\\\a_4=-3(78)-12=-234-12=-246\\\\\text{for}\ n=5\\\\a_5=-3a_{5-1}-12=-3a_4-12\\\\\text{substitute}\ a_4=-246\\\\a_5=-3(-246)-12=702-12=690

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