Question

Find the sum to n terms
i) 2+22+222+....
ii) 5+55+555+....

Answer 1

1. 246
2. 615

just trying.....

if wrong tell me

hope it's helpful

Answer 2

Answer:

$i)S=\dfrac{2}{81}(10^{n+1}-10-9n)$

$ii)S=\dfrac{5}{81}(10^{n+1}-10-9n)$

Step-by-step explanation:

Find the sum of n terms:

i)2+22+222+..........

Let S=2+22+222+...........upto n terms[Where S is the sum of the  series]

S=2(1+11+111+...............upto n terms)

Multiply and divide by 9 to the above term.

$S=\dfrac{2}{9}(9+99+999+.............upto\ n\ terms)$

$S=\dfrac{2}{9}[(10-1)+(10^2-1)+(10^3-1)+.........+upto\ n\ terms)$

$S=\dfrac{2}{9}[10+10^2+10^3+.............+upto\ n\ terms]-n$

Sum=1+1+1+1+1+.................upto n terms =n

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

Where number of term=n and ratio=10

General series of G.P. is:

$a+ar+ar^2+ar^3+..............ar^n-1$

$r=\dfrac{ar^n}{ar^{n-1}}$

Here r is the ratio and n is the number of term.

$S=\dfrac{2}{9}\dfrac{(10)(10^n-1)}{10-1}-n$

$S=\dfrac{2}{9}\dfrac{(10)(10^n-1)-9n}{9}$

$S=\dfrac{2}{81}(10^{n+1}-10-9n)$

Find the sum of n terms:

ii)5+55+555+..........

Let S=5+55+555+...........upto n terms [Where S is the sum of the  series]

S=5(1+11+111+...............upto n terms)

Multiply and divide by 9 to the above term.

$S=\dfrac{5}{9}(9+99+999+.............upto\ n\ terms)$

$S=\dfrac{5}{9}[(10-1)+(10^2-1)+(10^3-1)+.........+upto\ n\ terms)$

$S=\dfrac{5}{9}10+10^2+10^3+.............+upto\ n\ terms-n$

Sum=1+1+1+1+1+.................upto n terms =n

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

Where number of term=n and ratio=10

Here r is the ratio and n is the number of term.

$S=\dfrac{5}{9}\dfrac{(10)(10^n-1)}{10-1}-n$

$S=\dfrac{5}{9}\dfrac{(10)(10^n-1)-9n}{9}$

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