7+77+777+....(evaluate the summation of n-terms)?

Question

please give me step by step solution.

Answer 1

Every term is 10 times the prior, plus 7.

The terms are described as follows:
a(n) = 10 a(n-1) + 7

In closed form, each term is given by the formula:
a(n) = (10^n-1)7/9

In closed form, the summation of n terms is given as:
s(n) = (10^n-1) * (70/81) - (7/9)n

s(1) = 9*70/81 - 7/9 = 63/9 = 7
s(2) = 99*70/81 - 14/9 = (770-14)/9 = 756/9 = 84
s(3) = 999*70/81 - 21/9 = (7770-21)/9 = 7749/9 = 861
s(4) = 9999*70/81 - 28/9 = (77770-28)/9 = 77742/9 = 8638
s(5) = 99999*70/81 - 35/9 = (777770 - 35) = 777735/9 = 86415
etc.

Answer 2

For n equal to a positive integer (being the number of terms) the total is equal to 7*{ 1*10^(n-1) + 2*10^(n-2) + ... + (n-i)*10^i + ... + (n-1)*10^2 + n*10^0 }

there are a few steps, start by factoring out the 7. notice the terms become 1, 11, 111, 1111, etc. and the totals become 1, 12, 123, 1234, etc. then you will have to do the algebra.

Answer 3

The k-th term is equal to 7/9 * (10^k - 1).

The sum of n terms is

SUM 7/9 * (10^k - 1) = 7/9 * [10/9*(10^n - 1) - n]

Check: if n = 4, we find

7/9 * [10/9 * 9999 - 4] = 7/9 * 11106 = 8638,

which is indeed 7 + 77 + 777 + 7777

Answer 4

the sum to ‘n’ terms of a G.P.is [a(r^n-1)]/r-1
7+77+777+………………………………………………n terms).this now has to be rewritten into two components,one a G.P. and another an A.P.
=7(1+11+111+………………………………………….n terms)
=7/9(9+99+999+………………………………………………n terms).
=7/9[(10-1)+(100-1)+(1000-1)+………………………….n terms]
=7/9(10+100+1000+………………n terms)-7/9(1+1+1+………………….n terms)
=(7/9)*[10*(10^n-1)]/(10-1)-7/9n

Answer 5

700+70=770+70=840+7=847

Answer 6

sum from i=1 until i=N ((N-i+1)*7*10^(i-1))