1. ## sequence problem

7+77+777... n terms sum=??

we take 7/9 as common and get 9,99,999... then we break it in form of 10-1+10*10-1...

is there any other way to solve this sum

2. Originally Posted by prasum
7+77+777... n terms sum=??

we take 7/9 as common and get 9,99,999... then we break it in form of 10-1+10*10-1...

is there any other way to solve this sum
That seems to me the easiest way to solve, and I can't think of another way right now.

With some manipulation you get $f(n)=(\frac{7}{9})((\frac{10}{9})(10^n-1)-n)$.

3. The pattern is quite pretty early on.
7 = 7*1
+77 = 84 = 7*12
+777 = 861 = 7*123
+7777 = 8638 = 7*1234
+77777 = 86415 = 7*12345
etc.
Things become a little more awkward when you reach ten sevens.

4. ## sequence

s= 7+77+777+..

can we take 7 as common the we get

7(1+11+111...)

therefore we will get the sum of 1+11+111...

as

1*10^(n-1)+2*10^(n-2)...
this is ap gp series so we can solve it

but it is getting tedious

also i dont know whether this is right or wrong

5. Originally Posted by prasum
s= 7+77+777+..

can we take 7 as common the we get

7(1+11+111...)

therefore we will get the sum of 1+11+111...

as

1*10^(n-1)+2*10^(n-2)...
this is ap gp series so we can solve it

but it is getting tedious

also i dont know whether this is right or wrong

$\displaystyle f(n)=\left(\frac{7}{9}\right)\left(\sum_{i=1}^n(10 ^i-1)\right)$

$\displaystyle =\left(\frac{7}{9}\right)\left(\sum_{i=1}^n10^i-\sum_{i=1}^n1\right)$

$=(\frac{7}{9})((\frac{10}{9})(10^n-1)-n)$

Reducing the first sum in the final step isn't so hard when you notice the numbers are just

10,
110,
1110, ...

To get 1110 you can take 999 and multiply by (10/9).