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Question: The sum of \(\frac { 3 } { 1.2 }\) . \(\frac { 1 } { 2 }\) + \(\frac { 4 } { 2.3 }\) \(\left( \fr...

The sum of 31.2\frac { 3 } { 1.2 } . 12\frac { 1 } { 2 } + 42.3\frac { 4 } { 2.3 } (12)2\left( \frac { 1 } { 2 } \right) ^ { 2 }+ 53.4\frac { 5 } { 3.4 } . (12)3\left( \frac { 1 } { 2 } \right) ^ { 3 }+ … to n terms, is equal to-

A

1 –

B

1– 1n⋅2n−1\frac { 1 } { \mathrm { n } \cdot 2 ^ { \mathrm { n } - 1 } }

C

1 + 1(n+1)⋅2n\frac { 1 } { ( n + 1 ) \cdot 2 ^ { n } }

D

None of these

Answer

1 –

Explanation

Solution

nth term of the series tn = 3+(n−1)n(n+1)\frac { 3 + ( \mathrm { n } - 1 ) } { \mathrm { n } ( \mathrm { n } + 1 ) } (12)n\left( \frac { 1 } { 2 } \right) ^ { n }

=

tn = (2n−1n+1)\left( \frac { 2 } { n } - \frac { 1 } { n + 1 } \right)

\ Sn == =

=

= (12.21−13.22)\left( \frac { 1 } { 2.2 ^ { 1 } } - \frac { 1 } { 3.2 ^ { 2 } } \right)+(13.22−14.23)\left( \frac { 1 } { 3.2 ^ { 2 } } - \frac { 1 } { 4.2 ^ { 3 } } \right)

+ …..+ = 1 –