S= 3+33+... S= 3(1+11+...) S= 13(9+99+...) S= 13((10 1)+(100 1)+...) = 13(10+100+... n) = 13 (10(10^n 1)9 n ) =127[10^n+1 10 ...

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Summation by Sigma Method

3+33+333+.......

Question

$3 + 33 + 333 + . . . . . +$ n terms.

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Solution

$S = 3 + 33 + . . .$
$S = 3 (1 + 11 + . . .)$
$S = \frac{1}{3} (9 + 99 + . . .)$
$S = \frac{1}{3} ((10 - 1) + (100 - 1) + . . .)$
$= \frac{1}{3} (10 + 100 + . . . - n)$
$= \frac{1}{3} (\frac{10 (10^{n} - 1)}{9} - n)$
$= \frac{1}{27} [10^{n + 1} - 10 - 9 n]$

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3 + 33 + 333 + 3333 +

Question 19

State whether the given statements are true (T) or false (F).

3 : 33 = 33 : 333

P = \frac{3}{17} + \frac{33}{17^{2}} + \frac{333}{17^{3}} + \dots \infty

P

Question 19

State whether the given statements are true (T) or false (F).

3 : 33 = 33 : 333

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