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September 27, 2019

Sum based on Probability - ISI MMA 2018 Question 24

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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" _i="1" _address="0.0.0.1"]The sum of Infinity series                 \( 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + .......\) is [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27.4" _i="0" _address="0.1.0.0.0"]

Sample Questions(MMA) : 2018 Problem no 24

[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27.4" _i="1" _address="0.1.0.0.1" open="off"]Probability[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27.4" _i="2" _address="0.1.0.0.2" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27.4" _i="3" _address="0.1.0.0.3" open="off"]A first course in Probability by Sheldon Ross[/et_pb_accordion_item][/et_pb_accordion][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_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.29.2" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27.4" _i="1" _address="0.1.0.2.1"]Do you know about Arithmatics-Geometric series .(geometric series is the sum of a geometric sequence. Thus, with the series you just see if the relationship between the terms is arithmetic (each term increases or decreases by adding a constant to the previous term ) or geometric (each term is found by multiplying the previous term by a constant).[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.29.2" _i="2" _address="0.1.0.2.2"]\(S_n = [ 1+ \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} +...........]\) \(  \frac{S_n}{3} = [ \frac{1}{3} + \frac{2}{3^2} + \frac{6}{3^3} + \frac{10}{3^4} +............] \) \( \\ \)   Now use this two to apply arithmatic-geometric series .[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.29.2" _i="3" _address="0.1.0.2.3"]\([S_n - \frac{1}{3}S_n]\) = [ 1 +\(\frac{1}{3}\) + \(\frac{4}{3^2}\) + \(\frac{4}{3^3}\) + \(\frac{4}{3^4}\)+..........]  \( \Rightarrow \frac{2}{3}S_n\) = \(\frac{4}{3}[ 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + ..............]\) = \( \frac{4}{3}\) *\( \frac{1}{1-\frac{1}{3}}\)  \( \Rightarrow \frac{2}{3}S_n\) = \(\frac{4}{3} * \frac{3}{2}\) =2  \( \Rightarrow S_n = 3\)[/et_pb_tab][/et_pb_tabs][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_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" _i="3" _address="0.1.0.3"]

Watch the video

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Similar Problems

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