Question

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I need help number 21 Please!

Answer 1

you have been asking these from morning ...did you got your number 11

Answer 2

well it's a geometric series

Answer 3

I post this 10 hours ago, I can't find it

Answer 4

btw it's answer was -52

Answer 5

now here we have infinite series

Answer 6

yes I got -52, but this not thesame number I asking

Answer 7

yes, I did try very hard but I can answer corect

Answer 8

\[Sum_{G.P} = \frac{7^n -1}{6}\]

Answer 9

Let me give you formula

Answer 10

\[Sum_{G.P} = a \times \frac{r^n -1}{r -1}\]

Answer 11

a is the first term of the sequence ....
r is the ratio

Answer 12

the series is 1 , 7, 49 , .....7^n-1

analogous to a , ar ,a (r^2) ,......a r^n-1

Answer 13

my eqution is \[7^{k+1-1}=7^k\]\[\frac{1}{6}[7^{K}-1]=\frac{1}{6}[7^{k+1}-1]\]

Answer 14

I add\[\frac{1}{6}(7^k-1)+7^{k+1}=\frac{1}{6}(7^{k+1}-1]\]

Answer 15

this 2 equation not match, I spend many hours to solve

Answer 16

See Nth term is 7^n-1

so we have 1 + 7 + 49 + ...+ 7^n-1

\[\sum_{k=0}^{n -1} 7^k\]

Answer 17

\[(1 - r ) \sum_{0}^{n-1} = (1 - r) ( a + ar +a r^2 +...ar^n-1)\]

Answer 18

I hope you got that i missed 7^k

Answer 19

I did sept 1,2
stuck step 3, can't prove true for n=k+1
2 equation solve not math

Answer 20

a =1 and r = 7
then when we open right hand side

we have a - ar + ar -ar^2 +.....+ar^n-1 - ar^n-1 +ar^n

Answer 21

which gives lhs = a + ar^n

Answer 22

\[\sum_{k=0}^{n-1} 7^k = \frac{a(1 - r^n)}{1 -r}\]

Answer 23

sorry typo in my second last and third last replies it should be -ar^n not + ar^n

Answer 24

now enter a =1 and r = 7 and you will have your answer

Answer 25

salina did it help ?

Answer 26

yes , thank