mathematical induction

Question

7+77+777+.........+(777......n times)=

Answer 1

This is the same as 7(1+11+111+...). Note that the individual terms in parentheses are each a geometric progression (GP) with first term 1 and common ratio 10, i.e., 1+10+10²+10³+... so the sum in parentheses is the sum of a sequence of GPs.

For n≥0 the nth term is 7(10ⁿ⁺¹-1)/(10-1)=(7/9)(10ⁿ⁺¹-1).

For example, when n=0, the term is (7/9)(10-1)=7;

when n=3, the term is (7/9)(10⁴-1)=(7/9)(9999)=7777.

The sum of the GP terms is S_n:

S_n=(7/9)(10¹-1+10²-1+10³-1+...+10ⁿ⁺¹-1)=

(7/9)(10¹+10²+10³+...+10ⁿ⁺¹-(n+1))=

(70/9)(1+10+10²+...+10ⁿ-(n+1)/10)=

(70/9)[(10ⁿ⁺¹-1)/9-(n+1)/10)]=

(70/9)[10(10ⁿ⁺¹-1)-9(n+1)]/90=

(7/81)(10ⁿ⁺²-9n-19).

When n=0, S₀=(7/81)(100-19)=7;

when n=3, S₃=(7/81)(100000-27-19)=(7/81)(99954)=8638.

7+77+777+7777=8638. Note that S₃ is the sum of the first 4 terms because n starts at zero.

So the formula is correct: S_n=(7/81)(10ⁿ⁺²-9n-19) for integer n≥0. Under this formula, S_n means the sum of the first n+1 terms.

Note that the sum of the first 9 terms is S₈=7×123,456,789=864,197,523; S₇=7×12,345,678=86,419,746, etc.

However, S₉=7×1,234,567,900=8,641,975,300;

S₁₀=7×12,345,679,011=86,419,753,077.

The formula can be written S_n=(7/81)(10ⁿ⁺¹-9n-10) so S₁=(7/81)(81)=7 for integer n≥1. Under this formula S_n means the sum of the first n terms. S₁ is just the first term in this case.