Question

Find the sum of 5 + 55 + 555 +..... to n terms.

Answer 1

Find the sum of 5 + 55 + 555 +..... to n terms.

5 (1 + 11 + 111 +... to n terms)

= 5×9(1 + 11 + 111 .... to n terms) /9

= 5(9 + 99 + 999+.... to n terms) /9

= 5[(10-1) + (10²-1) + (10³-1)+.... n terms) / 9

= 5[(10 + 10² + 10³+.... - n)] /9

= 5/9 × [(10× 10ⁿ-1)/10-1 - n]

= 5/9 × [(10× 10ⁿ-1)/9 - n]

= 5/9 × [(10× 10ⁿ-1 - 9n)/9]

= 5/81 × (10^n+1 - 9n - 10).

Answer 2

$\mathsf{5+55+555+...n\:terms}\\\\\implies \mathsf{5(1+11+111+.... n\:terms)}\\\\\implies \mathsf{\dfrac{5}{9}(9+99+999+... n\:terms)}\\\\\implies \mathsf{\dfrac{5}{9}(10-1+100-1+... n\:terms)}\\\\\implies \mathsf{\dfrac{5}{9}(10+100+.. n\:terms-1-1... n\:terms)}\\\\\implies \mathsf{\dfrac{5}{9}(10(\dfrac{10^n-1}{10-1})-n)}$

$\implies \mathsf{\dfrac{5}{9}(10(\dfrac{10^n-1}{9})-n)}\\\\\implies \mathsf{\dfrac{5}{9}((\dfrac{10^{n+1}-10}{9})-n)}\\\\\implies \mathsf{\dfrac{5}{9}(\dfrac{10^{n+1}-10-9n}{9})}\\\\\implies \mathsf{\dfrac{5}{81}(10^{n+1}-10-9n)}$

EXPLANATION :-

The given series is not an A.P or a G.P .

So we have to make the given series a G.P .

We start by taking 5 as common for all the elements .

Then the given series becomes 5 ( 1 + 11 + ..... n terms ) .

Then divide the whole series by 9 to get :

5/9 ( 9 + 99 + ..... n terms )

Then split each term into a G.P .

9 can be written as 10 - 1 .

99 can be written as 100 - 1 .

So we converted the data into a G.P of 10 + 100 .... n terms .

We know that ( - 1 - 1 .... n terms ) = - n .

Use the formula for sum of n terms of G.P .

$\mathtt{S_n=a\dfrac{r^n-1}{r-1}}$

Then further simplification will give us the answer !