Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

$\rm \longrightarrow {\bigg( 1\dfrac{3}{5}\bigg) }^{2} +{\bigg( 2\dfrac{2}{5}\bigg) }^{2} + {\bigg( 3\dfrac{1}{5}\bigg) }^{2} + {4}^{2} + {\bigg( 4\dfrac{4}{45}\bigg) }^{2} ...$

$\rm \longrightarrow {\bigg( \dfrac{8}{5}\bigg) }^{2} +{\bigg( \dfrac{12}{5}\bigg) }^{2} + {\bigg( \dfrac{16}{5}\bigg) }^{2} + {4}^{2} + {\bigg( \dfrac{24}{5}\bigg) }^{2} ...$

$\rm \longrightarrow {\bigg(2 \times \dfrac{4}{5}\bigg) }^{2} +{\bigg(3 \times \dfrac{4}{5}\bigg) }^{2} + {\bigg(4 \times \dfrac{4}{5}\bigg) }^{2} + {\bigg( 5 \times \dfrac{4}{5} }\bigg)^{2} + {\bigg(6 \times \dfrac{4}{5}\bigg) }^{2} ...$

Taking out (4/5)² as common :-

$\rm \longrightarrow {\bigg( \dfrac{4}{5}\bigg) }^{2} \bigg( {2}^{2} + {3}^{2} + {4}^{2} + ...{11}^{2} \bigg)$

$\rm \longrightarrow {\bigg( \dfrac{4}{5}\bigg) }^{2} \bigg( {2}^{2} + {3}^{2} + {4}^{2} + ...{11}^{2} + {1}^{2} - {1}^{2} \bigg)$

$\rm \longrightarrow {\bigg( \dfrac{4}{5}\bigg) }^{2} \bigg( {1}^{2} + {2}^{2} + {3}^{2} + {4}^{2} + ...{11}^{2} - {1}^{2} \bigg)$

$\rm \longrightarrow {\bigg( \dfrac{16}{25}\bigg) } \bigg( \dfrac{11 \times (11 + 1) \times (22 + 1)}{6} - 1 \bigg)$

$\rm \longrightarrow {\bigg( \dfrac{16}{25}\bigg) } \bigg( \dfrac{11 \times 2 \times 23}{1} - 1 \bigg)$

$\rm \longrightarrow {\bigg( \dfrac{16}{25}\bigg) } \bigg( \dfrac{2 2 \times 23}{1} - 1 \bigg)$

$\rm \longrightarrow {\bigg( \dfrac{16}{25}\bigg) } \bigg( 506 - 1 \bigg)$

$\rm \longrightarrow \dfrac{16}{25} \times 505$

$\rm \longrightarrow \dfrac{16}{5} \times 101$

Sum of the series is given as (16/5)m.

$\rm \longrightarrow \dfrac{16}{5} \times 101 = \dfrac{16}{5} m$

$\rm \longrightarrow 101 =m$

Answer :-

Option (b) 101 is correct

Answer 2

Step-by-step explanation:

\rm \longrightarrow {\bigg( 1\dfrac{3}{5}\bigg) }^{2} +{\bigg( 2\dfrac{2}{5}\bigg) }^{2} + {\bigg( 3\dfrac{1}{5}\bigg) }^{2} + {4}^{2} + {\bigg( 4\dfrac{4}{45}\bigg) }^{2} ... ⟶(1

5

3

)

2

+(2

5

2

)

2

+(3

5

1

)

2

+4

2

+(4

45

4

)

2

...

\rm \longrightarrow {\bigg( \dfrac{8}{5}\bigg) }^{2} +{\bigg( \dfrac{12}{5}\bigg) }^{2} + {\bigg( \dfrac{16}{5}\bigg) }^{2} + {4}^{2} + {\bigg( \dfrac{24}{5}\bigg) }^{2} ...⟶(

5

8

)

2

+(

5

12

)

2

+(

5

16

)

2

+4

2

+(

5

24

)

2

...

\rm \longrightarrow {\bigg(2 \times \dfrac{4}{5}\bigg) }^{2} +{\bigg(3 \times \dfrac{4}{5}\bigg) }^{2} + {\bigg(4 \times \dfrac{4}{5}\bigg) }^{2} + {\bigg( 5 \times \dfrac{4}{5} }\bigg)^{2} + {\bigg(6 \times \dfrac{4}{5}\bigg) }^{2} ...⟶(2×

5

4

)

2

+(3×

5

4

)

2

+(4×

5

4

)

2

+(5×

5

4

)

2

+(6×

5

4

)

2

...

Taking out (4/5)² as common :-

\rm \longrightarrow {\bigg( \dfrac{4}{5}\bigg) }^{2} \bigg( {2}^{2} + {3}^{2} + {4}^{2} + ...{11}^{2} \bigg)⟶(

5

4

)

2

(2

2

+3

2

+4

2

+...11

2

)

\rm \longrightarrow {\bigg( \dfrac{4}{5}\bigg) }^{2} \bigg( {2}^{2} + {3}^{2} + {4}^{2} + ...{11}^{2} + {1}^{2} - {1}^{2} \bigg)⟶(

5

4

)

2

(2

2

+3

2

+4

2

+...11

2

+1

2

−1

2

)

\rm \longrightarrow {\bigg( \dfrac{4}{5}\bigg) }^{2} \bigg( {1}^{2} + {2}^{2} + {3}^{2} + {4}^{2} + ...{11}^{2} - {1}^{2} \bigg)⟶(

5

4

)

2

(1

2

+2

2

+3

2

+4

2

+...11

2

−1

2

)

\rm \longrightarrow {\bigg( \dfrac{16}{25}\bigg) } \bigg( \dfrac{11 \times (11 + 1) \times (22 + 1)}{6} - 1 \bigg)⟶(

25

16

)(

6

11×(11+1)×(22+1)

−1)

\rm \longrightarrow {\bigg( \dfrac{16}{25}\bigg) } \bigg( \dfrac{11 \times 2 \times 23}{1} - 1 \bigg)⟶(

25

16

)(

1

11×2×23

−1)

\rm \longrightarrow {\bigg( \dfrac{16}{25}\bigg) } \bigg( \dfrac{2 2 \times 23}{1} - 1 \bigg)⟶(

25

16

)(

1

22×23

−1)

\rm \longrightarrow {\bigg( \dfrac{16}{25}\bigg) } \bigg( 506 - 1 \bigg)⟶(

25

16

)(506−1)

\rm \longrightarrow \dfrac{16}{25} \times 505⟶

25

16

×505

\rm \longrightarrow \dfrac{16}{5} \times 101⟶

5

16

×101

Sum of the series is given as (16/5)m.

\rm \longrightarrow \dfrac{16}{5} \times 101 = \dfrac{16}{5} m⟶

5

16

×101=

5

16

m

\rm \longrightarrow 101 =m⟶101=m

Answer :-

Option (b) 101 is correct