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# Sum Of 1'S C.M.I UG-2019 Entrance

## Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"] Find the sum 1+111+11111+1111111+.....1....111(2k+1) ones

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Algebra

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3.5 out of 10

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challenges and trills of pre college mathematics

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Do you really need a hint? Try it first!

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can you some how manipulate it in geometric progression .

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$\frac{1}{9}(9+99+999+9999.........)$

now can u transfer it into G.P series

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ok let's see

$\frac{1}{9}[(10^1-1)+(10^2-1)+.........]$ upto 2k+1 terms

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

now can u see the G.P Series , with first term 10 and common ratio 10

use the formula ,$\frac{a(1-r^n)}{r-1}$

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so the final ans is  $\frac{1}{9}[\frac{10}{9}(10^{2k+1}-1)-n]$

## Connected Program at Cheenta

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.