GATE 2020 IN GA Q.8: solution and answer

GATE 2020 · INGA Q.8 · 2 marks · MCQ

The sum of the first $n$ terms in the sequence 8, 88, 888, 8888, ... is______.

(A)
$\frac{81}{80} (1 0^{n} - 1) + \frac{9}{8} n$
(B)
$\frac{81}{80} (1 0^{n} - 1) - \frac{9}{8} n$
(C)
$\frac{80}{81} (1 0^{n} - 1) + \frac{8}{9} n$
(D)
$\frac{80}{81} (1 0^{n} - 1) - \frac{8}{9} n$

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Answer and solution

ANSWER

The sum is $\frac{80}{81} (1 0^{n} - 1) - \frac{8}{9} n$ , which is option (D).

Official answer: DSource: Official answer key, GATE 2020 IN (IIT Delhi)

Take out the common factor 8, then write each repunit as a string of 9s divided by 9: $8 + 88 + 888 + \dots = 8 (1 + 11 + 111 + \dots) = \frac{8}{9} (9 + 99 + 999 + \dots)$ .
A string of $k$ nines is one less than a power of 10: $k 99 \dots 9 = 1 0^{k} - 1$ .
So the bracket is $k = 1 \sum n (1 0^{k} - 1) = (10 + 1 0^{2} + \dots + 1 0^{n}) - n$ .
Sum the GP with first term $a = 10$ and ratio $r = 10$ : $\frac{a ( r ^{n} - 1 )}{r - 1} = \frac{10 ( 1 0 ^{n} - 1 )}{9}$ .
Put the pieces together: $S_{n} = \frac{8}{9} [\frac{10 ( 1 0 ^{n} - 1 )}{9} - n] = \frac{80}{81} (1 0^{n} - 1) - \frac{8}{9} n$
Check with $n = 1$ : $\frac{80}{81} \cdot 9 - \frac{8}{9} = \frac{80 - 8}{9} = 8$ . Check with $n = 2$ : $\frac{80}{81} \cdot 99 - \frac{16}{9} = \frac{7920 - 144}{81} = 96 = 8 + 88$ . Both are correct.
This matches option (D).

KEY IDEA

Write each term as $\frac{8}{9} (1 0^{k} - 1)$ , since a string of $k$ eights is $\frac{8}{9}$ of a string of $k$ nines. The sum then splits into a geometric series in powers of 10 minus $n$ copies of 1. The fastest way to decide between options like these is to put in $n = 1$ and $n = 2$ and check against 8 and 96.

WATCH OUT FOR

Forgetting to subtract the $n$ (the " $- 1$ " inside each $1 0^{k} - 1$ ) gives a plus sign and points to (C). Starting the GP at $1 0^{0} = 1$ instead of $1 0^{1} = 10$ loses the factor 10 and gives a coefficient of $\frac{8}{81}$ instead of $\frac{80}{81}$ . Mixing up $\frac{8}{9}$ and $\frac{9}{8}$ leads to (A) or (B). Putting in $n = 1$ (the sum must be 8) rules out every wrong option at once.

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GATE 2020 IN GA Q.8: solution and answer

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