How to Find the Sum of 7 + 77 + 777 + … Up to n Terms

A general series-summation technique question, not tied to a specific exam paper.

Problem

Find the sum of the series

$$7 + 77 + 777 + \cdots \text{ up to } n \text{ terms}$$

Key insight. This series is neither an A.P. nor a G.P. — the gap between consecutive terms (70, then 700, then 7000, …) is itself growing like a G.P. The way in is to first find a clean formula for the general $n$th term, then sum that formula — rather than trying to sum the original series directly.

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Approach

Every term in this series is a string of repeated 7s: $7, 77, 777, \ldots$ — numbers built from $n$ sevens are called repunit-like numbers, and they have a standard closed form. Once the $n$th term is written that way, summing the whole series just means summing a geometric series (for the $10^n$ part) and a simple arithmetic count (for the constant part) separately.

Solution

Step 1 — Find a formula for the nth term

The $n$th term $T_n$ is a number made of $n$ sevens: for example, $T_1=7$, $T_2=77$, $T_3=777$. Each of these can be written as 7 times a string of $n$ ones — a repunit:

$$T_n = 7 \times \underbrace{11\ldots1}_{n \text{ ones}} = \frac{7}{9}\big(10^n – 1\big)$$

This works because $\underbrace{11\ldots1}_{n} = \dfrac{10^n-1}{9}$ (a standard identity: $99\ldots9 = 10^n-1$, and dividing by 9 gives all 1s).

Checking it: for $n=1$, $T_1 = \frac{7}{9}(10-1) = \frac{7}{9}(9) = 7$ ✓. For $n=2$, $T_2 = \frac{7}{9}(100-1) = \frac{7}{9}(99) = 77$ ✓.

Step 2 — Sum the nth-term formula from 1 to n

$$S_n = \sum_{k=1}^{n} T_k = \frac{7}{9}\sum_{k=1}^{n}\big(10^k – 1\big) = \frac{7}{9}\left[\sum_{k=1}^{n}10^k – \sum_{k=1}^{n}1\right] = \frac{7}{9}\left[\sum_{k=1}^{n}10^k – n\right]$$

Step 3 — Sum the geometric part

$\sum_{k=1}^{n} 10^k = 10 + 10^2 + \cdots + 10^n$ is a G.P. with first term 10, common ratio 10, and $n$ terms:

$$\sum_{k=1}^{n}10^k = \frac{10(10^n-1)}{10-1} = \frac{10^{n+1}-10}{9}$$

Step 4 — Combine everything

$$S_n = \frac{7}{9}\left[\frac{10^{n+1}-10}{9} – n\right] = \frac{7(10^{n+1}-10)}{81} – \frac{7n}{9}$$

Writing this with a single common denominator of 81:

$$S_n = \frac{7}{81}\Big(10^{n+1} – 9n – 10\Big)$$

Answer

Common mistakes

• Trying to spot a common difference or common ratio directly in the original series. The series itself is neither an A.P. nor a G.P. — only once it’s rewritten via the repunit identity does a clean G.P. appear.
• Forgetting the $-n$ term from $\sum 1$. Splitting $\sum(10^k – 1)$ into $\sum 10^k – \sum 1$ is easy to get right, but it’s equally easy to accidentally drop the second sum when combining terms at the end.

Practise next

• Find the sum of $5 + 55 + 555 + \cdots$ up to $n$ terms, using the same repunit-formula approach.