7 + 77 + 777 + ....+ 777...(n digits)...7 = 7/81 (10^(n+1) - 9n - 10)

Prove by mathematical induction.

Let P(n) be the statement given by

P(n) : 7 + 77 + 777 + ..... + 777...(n digits)...7 = 7/81 (10ⁿ⁺¹ - 9n - 10)

Step 1. For n = 1, we have

L.H.S. =7, R.H.S. =7/81(10¹⁺¹ - 9.1 - 10) = 7

= L.H.S. = R.H.S.

Thus, P(1) is true.

Step 2.For n=k, assume that P(k) is true, i.e.

7 + 77 + 777 + ..... + 777....(k digits)...7 = 7/81(10^k+1 - 9k - 10) ............(1)

Step 3. For n = k + 1, we have to show that P(k + 1) is true, using step 2.

i.e. 7 + 77 + 777 + ..... + 777...{(k + 1) digits}...7 = 7/81 {10^{k + 1 + 1} - 9(k + 1) - 10}

[777......{(k + 1) digits}...7=7.{111....[(k + 1)digits]...1}

=(7/9){999...[(k + 1) digits]...9}

=(7/9)(10^{k + 1} - 1)

]

Now, 7 + 77 + 777 + ..... + 777...(k digits)...7 + 777...{(k + 1) digits}...7

=7/81 {10^{k + 1} - 9k - 10} + 777...[(k + 1) digits]...7 [From (1)]

=7/81 {10^{k + 1} - 9k - 10} + (7/9)(10^{k + 1} - 1)

=7/81 (10^{k + 1} - 9k - 10 + 9.10^{k + 1} - 9)

=7/81 [(1 + 9)10^{k + 1} - 9k - 9 - 10]

=7/81 [10^{(k + 1) + 1} - 9(k + 1) - 10]

Therefore, P(K + 1) is true.

Thus, P(k) is true = P(k + 1) is true.

Hence, by the Principle of Mathematical Induction P(n) is true for all n belongs to N.