What is the sum of the first 20 term of the sequence 0.7, 0.77, 0.777,

why is c wrong i thought i nailed this question and damn sure that the option i marked (c) is correct

Solution:

Recall that 7/9 = 0.777…., 7/90 = 0.0777…, 7/900 - 0.00777… and so on. So we have:

1st term = 0.7 = 0.777… - 0.0777… = 7/9 - 7/90 = 7/9 - 7/9 x 10^-1

2nd term = 0.77 = 0.777… - 0.00777… = 7/9 - 7/900 = 7/9 - 7/9 x 10^-2

3nd term = 0.777 = 0.777… - 0.000777… = 7/9 - 7/9000 = 7/9 - 7/9 x 10^-3

…

20th term = 7/9 - 7/9 x 10^-20

Therefore, the sum of the first 20 terms is:

7/9 x 20 - 7/9 x (10^-1 + 10^-2 + 10^-3 + … + 10^-20)

= 7/9 x [20 - (10^-1 + 10^-2 + 10^-3 + … + 10^-20)]

= 7/9 x (20 - 0.11...11) (Note: 20 ones to the right of the decimal point)

= 7/81 x 9(20 - 0.11...11)

= 7/81 x (180 - 9(0.11...11))

= 7/81 x (180 - 0.99...99) (Note: 20 nines to the right of the decimal point)
= 7/81 x (180 - 1 + 0.00...01) (Note: 19 zeros between the decimal point and the rightmost 1)

= 7/81 x (179 + 10^-20)

Alternate Solution:

Let n = 0.7 + 0.77 + 0.777 + … + 0.77...77 (Note: The last term has 20 sevens to the right of the decimal point.)

So we are looking for the value of n. Multiplying the equation by 10, we have:

10n = 7 + 7.7 + 7.77 + … + 7.77...77 (Note: The last term has 20 sevens to the right of the decimal point.)

Subtracting the first equation from the second, we have:

9n = 7 + 7 + 7 + … + 7 - 0.77...77 (Note: The right hand side of the equation has 21 terms, the first 20 terms are 7 and the last one is the last term from the first equation.)

We can rewrite this as:

9n = 140 - 7(0.11...11) (Note: 20 ones to the right of the decimal point)

9n = 7(20 - 0.11...11)

n = 7/9 x (20 - 0.11...11) (Note: The right hand side is the third line of the long process in the first solution at the end. Therefore, we will get the eventual answer of 7/81 x (179 + 10^-20) also.)

Answer: C

0.7+0.77+0.777.....
= 7[(1/10)+(11/100)+(111/1000).....]

= 7/9[(9/10)+(99/100)+(999/1000)..]

= 7/9[(1-1/10)+(1-1/100)+(1-1/1000)......]

= 7/9[20- {(1/10)+(1/100)+(1/1000)+....]

=7/9[20- {(1/10)(1-(1/10)^20)/9/10}

= 7/9[20- (1/9) +(1/9)*10^{-20}]

= 7/9[(179/9) + (1/9)*10^{-20}]

= 7/81[179+10^{-20}]
Hence IMO C

Asked: What is the sum of the first 20 term of the sequence 0.7, 0.77, 0.777, … ?

\(S = .7 + .77 + .777 + .7777 + ..... + .77777---- 20 times 7 ----\)
\(= 7 (.1 + .11 + .111 + ... + .1111 -- 20 times 1 ---)\)
\(= \frac{7}{9} (.9 + .99 + .999 + .... + .9999---20 times 9--)\)
\(= \frac{7}{9} (1-.1 + 1-.01 + 1-.001 + .... + 1 - 10^{-20})\)
\(= \frac{7}{9} ( 20 - (.1 + .01 + .001 + ..... + 10^{-20}))\)
\(= \frac{7}{9} (20 - .1(1- 10^{-20})/(1-.1) ) \)
\(= \frac{7}{9} (20 - 1/9 (1- 10^{-20}))\)
\(= \frac{7}{81} (180 - (1-10^{-20})) \)
\(= \frac{7}{81} (179 + 10^{-20})\)

IMO C­