Question

Find the sum of the sequence 3 + 33 + 333 +.......+n.

Answer 1

Let the sum of the given sequence be $S_{n}$

So, $S_{n}$ = 3 + 33 + 333 + .... + n

$S_{n}$ = 3( 1 + 11 + 111 + .... upto\:n\:terms )

Divide and multiply by 9.

$S_{n}=\dfrac{3}{9}\bigg( 9 + 99 + 999 + ...\mathsf{upto\:n\:terms}\:\bigg)$

$S_{n} = \dfrac{1}{3}\bigg[ ( 10 - 1 ) + ( 100 - 1 ) + ( 100 - 1 ) + .... + \mathsf{upto\:n\:terms}\bigg]$

$S_{n} = \dfrac{1}{3}\bigg[10 - 1 + 100 - 1 + 1000 - 1 + .... + \mathsf{upto\:n \:terms} \bigg]$

$S_{n} = \dfrac{1}{3} \bigg[ 10 + 100 + 1000 + ... + \mathsf{upto\:n\:terms - 1 - 1 - 1 - ... upto\:n\:terms} \bigg]$

$S_{n} = \dfrac{1}{3}\bigg[ 10 + 100 + 1000 + .. + \mathsf{upto\:n\:terms )} - ( 1 + 1 + 1 + ... +\mathsf{ upto\:n\:terms} ) \bigg]$

Now, there are two progressions.One is geometric progression and other is arithmetic progression.

In Geometric progression,( 10 + 100 + 1000 + .... + upto n terms )

First term = a = 10

Common ratio = $\dfrac{1000}{100} = \dfrac{100}{10} = 10$

last term = $a_{l}$ = n th term

From the properties of geometric progressions,

Sₓ = $\dfrac{a(r^{n} -1)}{r-1}$

In Arithmetic progression,( 1 + 1 + 1 + ... + upto n terms )

First term = a = 1

Last term = $a_{l}$ = 1

$\therefore S_{n} = \dfrac{1}{3} \bigg[10 \times \bigg( \dfrac{10^n - 1 }{10-1} \bigg) - \dfrac{n}{2}\bigg( 1 + 1 \bigg) \bigg]$

$S_{n} = \dfrac{1}{3} \bigg[ \dfrac{10^{n+1} - 10 - 9n}{9}\bigg]$

$S_{n} = \dfrac{1}{27} \bigg[ \dfrac{10^{n+1} - 10 - 9n}{9} \bigg]$

therefore the sum of 3 + 33 + 333 + ..... + n terms is $\dfrac{1}{27} \bigg[ \dfrac{10^{n+1} - 10 - 9n}{9} \bigg]$

Answer 2

Given Sequence is 3 + 33 + 333 + ..... + n.

⇒ 3(1 + 11 + 111 + .... +n terms)

Take 3 as common factor and multiply & divide by 9, we get

$=> \frac{3}{9}[9 + 99 + 999 + ...... + n]$

$=> \frac{3}{9}[(10 - 1) + (10^2 - 1) + (10^3 - 1) + .... + n]$

$=> \frac{3}{9}[(10 + 10^2 + 10^3 + .... + 10^n) - (1 + 1 + 1 + ..... + 1)]$

$=> \frac{3}{9}[10(\frac{10^n - 1}{10 - 1}) - n]$

$=> \frac{3}{9}[ \frac{10}{9}(10^n - 1) - n]$

$=> \frac{3}{81}[(10^{n + 1} - 10 - 9n)]$

$=> \boxed{\frac{1}{27}[10^{n + 1} - 9n - 10]}$

Hope it helps!