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I have this question,

Prove,

$7 + 77 + 777 +7777 + 77...$n digits..$77 = 7/81[(10^n × 10) - 9n - 10]$

By induction.

Now since this question was given in the exercise that involves proving various statements by strong induction, this one is to be done by using strong induction.

Although I did it using the "weak" induction, I'm not quite sure how can this be, if possibly, done by strong induction?

Here is my assumption that let $P(n) : 7+ 77 + 777 +7777 + 77...$n digits..$77$ $= 7/81[(10^n × 10) - 9n - 10]$

For base cases, (n is supposed to be natural number in question, by the way)

For $P(1)$, each side = $7$

For $P(2)$, each side = $84$

So, now assume, $P(n)$ is true for any $i$ such that, $0 ≤ i ≤ k$ or, for all values of $n$ upto $k$, $P(n)$ is true. Then we have to prove that $P(k+1)$ is true.

So, $7+ 77 + 777 +7777 + 77...$k+1 digits..$77$ $= 7/81[((10^k × 10)× 10) - 9(k + 1) - 10]$ Is what we have to prove.

What do I do next? How do I prove the above statement making use of my inductive hypothesis?

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    $\begingroup$ I would start probably by writing the problem down using $\Sigma$ notation for the sums $\endgroup$ – Mason May 23 at 17:33
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    $\begingroup$ Hint: $777... n$ times $ = 7 \times (1111.... n \text{ times }) = 7 \times (10^n + 10^{n-1} + \dots 10^0) $. Does the last sum look familiar? $\endgroup$ – sudeep5221 May 23 at 17:38
  • $\begingroup$ @sudeep5221 I tried that way, but I am unable to use strong induction going that way. Would you elaborate, please? $\endgroup$ – user231094 May 24 at 0:14
  • $\begingroup$ @user231094 I am sorry I missed reading that you had got the answer by weak induction. My bad. As for strong induction, I can't think of any method that particularly leverages the strong induction hypothesis. To me it seems that weak induction is sufficient and perhaps the only induction based approach. $\endgroup$ – sudeep5221 May 24 at 2:39
  • $\begingroup$ The strong induction hypothesis implies the weak hypothesis, so if you can prove something by the weak hypothesis, that automatically also proves it by the strong hypothesis. The strong hypothesis gives you more material to work with. It does not require you to make use of each and every element of that material. $\endgroup$ – Paul Sinclair May 24 at 3:44

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