find the general term for summation of series given by 7+77+777+7777+77777+.................?

Answer 1

This is also 7 ∑ (1 + 11 + 111 + ... ).

Notice each of these numbers inside the term can be written in the form 10^0 + 10^1 + 10^2 + .... + 10^k. This is a geometric series, so its sum is (1 - 10^(k+1)) / (1-10), or
(10^(k+1) - 1)/9. So the series now becomes

7 ∑ [ (10^(1) - 1)/9 + (10^(2) - 1)/9 + (10^(3) - 1)/9 + ... ] =
(7/9) ∑ [ (10^(1) - 1) + (10^(2) - 1) + (10^(3) - 1) + ... ] =
(7/9)( [ ∑ 10^(1) + 10^(2) + 10^(3) + ... + 10^n ] - n )=
(7/9)( 10 [ ∑ 10^(0) + 10^(2) + 10^(3) + ... + 10^(n-1) ] - n )=
(7/9)( 10 [ (1 - 10^n) / (1-10) ] - n )=
(7/9)[ (10/9)(10^n - 1) - n ]

This is a general forumla for the sum, where "n" is the number of terms you have in the original problem.

Answer 2

This may be written as the double sum:

[j=1, n]∑[k=0, j-1]∑7*10^k

The inner term is a geometric series, and you will recall that the sum of a geometric series with terms a_1, a_2... a_n is simply (a_1-a_(n+1))/(1-r), where r is the common ratio. Thus, the inner series sums to:

[j=1, n]∑(7-7*10^j)/(1-10)

Or being simplified:

7/9 [j=1, n]∑(10^j-1)

Breaking this into two sums:

7/9 [j=1, n]∑10^j - 7/9 [j=1, n]∑1

Evaluating the geometric series and the constant sum:

7/9 (10-10^(n+1))/(-9) - 7n/9

Or being simplified:

(7*10^(n+1) - 63n - 70)/81

And we are done.

Answer 3

progressive convergence