[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]
[/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
[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.3.4"]C.M.I UG-2019 entrance exam
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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 \(\frac{1}{9}[\frac{10}{9}(10^{2k+1}-1)-n]\)
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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.
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