If <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> <mtext> </mtext> <mtext> </mtext> <mo>.</mo> <mtext> </mtext> <mtext> </mtext> <msup> <mrow> <mn>3</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msup> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mtext> </mtext> <mtext> </mtext> <mo>.</mo> <mtext> </mtext> <mtext> </mtext> <msup> <mrow> <mn>3</mn> </mrow> <mrow> <mn>9</mn> </mrow> </msup> </mrow> </mfrac> <mo>+</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msup> <mo>.</mo> <mn>3</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>K</mi> </mrow> <mrow> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msup> <mtext> </mtext> <mtext> </mtext> <mo>.</mo> <mtext> </mtext> <mtext> </mtext> <msup> <mrow> <mn>3</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msup> </mrow> </mfrac> <mo>,</mo> </mrow> </math> then the remainder when K is divided by 6 is:

t n = 1 2 n . 3 1 1 − n = ( 1 3 ) 1 1 . ( 3 2 ) n S 1 0 = ( 1 3 ) 1 1 . ( 3 2 ) . ( ( 3 2 ) 1 0 − 1 ) 3 2 − 1 = 1 3 1 0 . 3 1 0 − 2 1 0 2 1 0 = 1 + 1 0 C 1 . 2 1 + 1 0 C 2 . 2 2 + . . . . + 1 0 C 9 . 2 9 k = 21 + 6m ® R = 3

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Maths NCERT Exemplar Solutions Class 11th Chapter Two Maths Class 11th Maths Binomial Theorem

If $\frac{1}{2 . 3^{10}} + \frac{1}{2^{2} . 3^{9}} + . . . . + \frac{1}{2^{10} . 3} = \frac{K}{2^{10} . 3^{10}},$ then the remainder when K is divided by 6 is:

Option 1 -

Option 2 -

Option 3 -

Option 4 -

2 Views | Posted 3 weeks ago

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Answered by

Raj Pandey | Contributor-Level 9

3 weeks ago

Correct Option - 3

Detailed Solution:

$t_{n} = \frac{1}{2^{n} . 3^{11 - n}} = {(\frac{1}{3})}^{11} . {(\frac{3}{2})}^{n}$

$S_{10} = {(\frac{1}{3})}^{11} . \frac{(\frac{3}{2}) . ({(\frac{3}{2})}^{10} - 1)}{\frac{3}{2} - 1}$

$= \frac{1}{3^{10}} . \frac{3^{10} - 2^{10}}{2^{10}}$

$= 1 +^{10} C_{1} . 2^{1} +^{10} C_{2} . 2^{2} + . . . . +^{10} C_{9} . 2^{9}$

k = 21 + 6m ® R = 3

<math> <mrow> <msub> <mrow> <mi>t</mi> </mrow> <mrow> <mi>n</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mi>n</mi> </mrow> </msup> <mo>.</mo> <msup> <mrow> <mn>3</mn> </mrow> <mrow> <mn>1</mn> <mn>1</mn> <mo>−</mo> <mi>n</mi> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>1</mn> <mn>1</mn> </mrow> </msup> <mo>.</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mi>n</mi> </mrow> </msup> </mrow> </math> <math> <mrow> <msub> <mrow> <mi>S</mi> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msub> <mo>=</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>1</mn> <mn>1</mn> </mrow> </msup> <mo>.</mo> <mfrac> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>.</mo> <mrow> <mo> (</mo> <mrow> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msup> <mo>−</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>−</mo> <mn>1</mn> </mrow> </mfrac> </mrow> </math> <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <msup> <mrow> <mn>3</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msup> </mrow> </mfrac> <mo>.</mo> <mfrac> <mrow> <msup> <mrow> <mn>3</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msup> <mo>−</mo> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msup> </mrow> <mrow> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math> <math> <mrow> <mo>=</mo> <mn>1</mn> <msup> <mrow> <mo>+</mo> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msup> <msub> <mrow> <mi>C</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>.</mo> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mn>1</mn> </mrow> </msup> <msup> <mrow> <mo>+</mo> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msup> <msub> <mrow> <mi>C</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>.</mo> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <msup> <mrow> <mo>+</mo> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msup> <msub> <mrow> <mi>C</mi> </mrow> <mrow> <mn>9</mn> </mrow> </msub> <mo>.</mo> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mn>9</mn> </mrow> </msup> </mrow> </math> k = 21 + 6m ® R = 3

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If $\frac{1}{2 . 3^{10}} + \frac{1}{2^{2} . 3^{9}} + . . . . + \frac{1}{2^{10} . 3} = \frac{K}{2^{10} . 3^{10}},$ then the remainder when K is divided by 6 is:

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If 1 2 . 3 1 0 + 1 2 2 . 3 9 + . . . . + 1 2 1 0 . 3 = K 2 1 0 . 3 1 0 , then the remainder when K is divided by 6 is:

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If $\frac{1}{2 . 3^{10}} + \frac{1}{2^{2} . 3^{9}} + . . . . + \frac{1}{2^{10} . 3} = \frac{K}{2^{10} . 3^{10}},$ then the remainder when K is divided by 6 is: