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Find the sum to $\mathsf{n}$ terms of the series $\mathsf{ 1 \: + \: \dfrac{3}{2} \: + \: \dfrac{5}{4} \: + \: \dfrac{7}{8}+.....}$.

Answer 1

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$





$\underline{\mathsf{Given \: Series \longrightarrow 1 \: + \: \dfrac{3}{2} \: + \: \dfrac{5}{4} \: + \: \dfrac{7}{8} \: + \: ..... }}$





$\textsf{This series is an Arithmetico - Geometric series ,} \\ \textsf{ in this type of sequence each term is product of } \\ \textsf{corresponding A.P. and G.P.}$





$\textsf{The above series can be expressed as this : } \\ \\ \sf = 1 \: \times \: \dfrac{1}{1} \: + \: \{ 1 \: + \: (2 \: - \: 1)2 \} \dfrac{1}{1 \: \times \: 2} \: + \: \{1 \: + \: (3 \: - \: 1)2 \} \dfrac{1}{1 \: \times \: {2}^{2} } \: + \: \{1 \: + \: (4 \: - \: 1)2 \} \dfrac{1}{1 \: \times \: {2}^{3} } \: + \: .....$




$\underline{\textsf{For A.P.,}} \\ \\ \sf \implies First \: term \: = \: 1 \\ \\ \sf \implies Common \: difference \: = \: 2 \\ \\ \textsf{Now,} \\ \\ \sf \implies T_n \: = \: a \: + \: (n \: - \: 1)d \\ \\ \sf \implies T_n \: = \: 1 \: + \: ( n \: - \: 1)2 \\ \\ \sf \implies T_n \: = \: 1 \: + \: 2n \: - \: 2 \\ \\ \sf\: \: \therefore \: \: T_n \: = \: 2n \: - \: 1$





$\underline{\textsf{For G.P.,}} \\ \\ \sf \implies First \: term \: = \: 1 \\ \\ \sf \implies Common \: ratio \: = \: \dfrac{1}{2} \\ \\ \textsf{Now,} \\ \\ \sf \implies T_n \: = \: ar^{( n \: - \: 1)} \\ \\ \sf \implies T_n \: = \: 1 \: \times \: \left(\dfrac{1}{2} \right)^{ ( n \: - \: 1)} \\ \\ \sf\: \: \therefore \: \: T_n \: = \: \dfrac{1}{ {2}^{(n \: - \: 1)} }$





$\underline{\textsf{Now,}} \\ \\ \sf \implies T_n \: of \: A.G.P \: = \: \dfrac{2n \: - \: 1}{2^{(n \: - \: 1 )}}$




$\textsf{Let,} \\ \\ \sf \implies S_n \: = \: 1 \: + \: \dfrac{3}{2} \: + \: \dfrac{5}{4} \: + \: \dfrac{7}{8} \: + \: ..... \: + \: \dfrac{2n \: - \: 1}{2^{(n \: - \: 1)} } \qquad...(1)$





$\textsf{Multiply both sides by common ratio , } \\ \\ \sf \implies \dfrac{1}{2} S_n \: = \dfrac{1}{2} \left\{\: 1 \: + \: \dfrac{3}{2} \: + \: \dfrac{5}{4} \: + \: \dfrac{7}{8} \: + ..... + \dfrac{2n \: - \: 1}{2^{(n \: - \: 1)} } \right\} \\ \\ \\ \sf \implies\dfrac{1}{2} S_n = \dfrac{1}{2} \: + \: \dfrac{3}{4} + \dfrac{5}{8} + \dfrac{7}{16} + ... + \dfrac{2n \: - \: 3}{ {2}^{(n \: - \: 1)} } \: + \: \dfrac{2n \: - \: 1}{ {2}^{n} } \qquad...(2)$





$\underline{\textsf{Subtract (2) from (1),}} \\ \\ \sf \implies S_n = 1 + \dfrac{3}{2} + \dfrac{5}{4} + \dfrac{7}{8} + ... + \dfrac{2n \: - \: 1}{2^{(n \: - \: 1)} } \\ \\ \\ \sf \implies\dfrac{1}{2} S_n = \dfrac{1}{2} \: + \: \dfrac{3}{4} \: + \: \dfrac{5}{8} \: + \: ... \: + \: \dfrac{2n \: - \: 3}{ {2}^{(n \: - \: 1)} } \: + \: \dfrac{2n \: - \: 1}{ {2}^{n} } \\ \\ \textsf{On Subtraction , we get : } \\ \\ \sf \implies \dfrac{1}{2} S_n \: = \: 1 \: + \: \dfrac{2}{2} \: + \: \dfrac{2}{4} \: + \: \dfrac{2}{8} \: + \: ... \: + \: \dfrac{2}{ {2}^{(n \: - \: 1)} } \: - \: \dfrac{2n \: - \: 1}{ {2}^{n} }$




$\sf \implies \dfrac{1}{2} S_n \: = \: 1 + 2\left\{ \underbrace{\dfrac{1}{2} \: + \: \dfrac{1}{4} + \dfrac{1}{8} + ... + \dfrac{1}{ {2}^{(n \: - \: 1)} }}_{This \: Part \: Forms \: an \: G.P.} \right \} \: - \: \dfrac{2n \: - \: 1}{ {2}^{n} }$





$\underline{\textsf{For this G.P.,}} \\ \\ \sf \implies First \: term \: = \: \dfrac{1}{2} \\ \\ \\ \sf \implies Common \: difference = \dfrac{1}{2} \\ \\ \sf \implies Last \: term = \dfrac{1}{2^{(n \: - \: 1)}}$




$\textsf{Let total number of terms is N.} \\ \\ \sf \implies T_N \: = \: ar^{(N \: - \: 1)} \\ \\ \sf \implies \dfrac{1}{2^{(n \: - \: 1)} } = \dfrac{1}{2} \: \times \: \left \{\dfrac{1}{2} \right \}^{(N \: - \: 1)} \\ \\ \sf \implies \dfrac{1}{ {2}^{(n \: - \: 1)} } = \dfrac{1}{2 ^{N }}$





$\textsf{On Comparison , } \\ \\ \sf \implies N \: = \: n \: - \: 1 \\ \\ \underline{\textsf{Now,}} \\ \\ \sf \implies \dfrac{1}{2}S_n \: = \: 1 \: + 2 \left \{ \dfrac{ \dfrac{1}{2} \left(1 \: - \: \dfrac{1}{2 {}^{n \: - \: 1} } \right)}{1 \: - \: \dfrac{1}{2} } \right \} \: - \: \dfrac{2n \: - \: 1}{ {2}^{n} } \\ \\ \\ \sf \implies \dfrac{1}{2}S_n \: = \: 1 \: + 2 \left \{ \dfrac{ \cancel{\dfrac{1}{2}} \left(1 \: - \: \dfrac{1}{2 {}^{n \: - \: 1} } \right)}{ \cancel{ \dfrac{1}{2} } } \right \} \: - \: \dfrac{2n \: - \: 1}{ {2}^{n} }$





$\sf \implies \dfrac{1}{2}S_n \: = \: 1 \: + \: 2\left( 1 \: - \: \dfrac{1}{2^{n \: - \: 1}} \right) \: - \: \dfrac{2n \: - \: 1}{2^n} \\ \\ \\ \sf \implies \dfrac{1}{2}S_n \: = \: 1 \: + \: 2 \: - \: \dfrac{2}{2^{n \: - \: 1}} \: - \: \dfrac{2n \: - \: 1}{2^n}$




$\sf \implies \dfrac{1}{2}S_n \: = \: 3 \: - \: \left( \dfrac{4}{2^n} \: + \: \dfrac{2n \: - \: 1}{2^n} \right) \\ \\ \\ \sf \implies \dfrac{1}{2}S_n \: = \: 3 \: - \: \left( \dfrac{4 \: + \: 2n \: - \: 1}{2^n} \right) \\ \\ \sf \implies \dfrac{1}{2}S_n \: = \: 3 \: - \: \left( \dfrac{ \: 2n \: + \: 3}{2^n} \right)$




$\sf \implies S_n \: =\: 2 \left( 3 \: - \: \dfrac{2n \: + \: 3}{2^n} \right) \\ \\ \sf \therefore S_n \: = \: 6 \: - \: \dfrac{2n \: + \: 3}{2^{(n \: - \: 1)} }$

Answer 2

$Given \: sequence :1 + \dfrac{3}{2} + \dfrac{5}{4} + \dfrac{7}{8} + ....upto \: n \: terms$


On observing the given sequence, we get : -

All the numerator are in arithmetic progression, whereas all the denominators are in geometric progressions. There are two progressions ( A.P. and G.P. ). These types of sequences are the part of arithmetico - geometric progressions( AGP ).



Note : For now, don't do anything with 1, solving the remaining part:



Now, there are n - 1 terms.

Given arithmetic progression ( denominator ) : 3 , 5 , 7 ..... upto n - 1 terms

Given geometric progression ( numerator ) : 2 , 4 ( or 2² ) , 8 ( or 2³ )..... upto n - 1 terms.


We know say that the n th term of the sequence should be $\dfrac{(n-1)\: th\:term\: of\:AP}{(n-1)\: th\: term\: of\: GP}$.


From the properties of arithmetic & geometric progressions, we know : -

• n - 1 th term of AP = a + { ( n - 1 ) - 1 }d = a + ( n - 2 )d

• n - 1 th term of GP = $ar^{(n-1) -1}=ar^{n-2}$


In the question :

First term of the AP[ excluding 1 ] : 3

Common Difference( b/w APs ) : 5 - 3 = 2

first term of GP : 2

Common ratio( b/w GPs ) : 2 / 4 = 1 / 2, but while solving for last term, we will write as 2 with the required power.


So,
Last term of this series[ excluding 1 ]:$\dfrac{3+(n-2)2}{2(2)^{n-2}}=\dfrac{2n-1}{2^{n-1}}$



Now,
$Sequence \: is:1 + \dfrac{3}{2} + \dfrac{5}{4} + \dfrac{7}{8} + ..... + \dfrac{2n-1}{2^{n - 1}}$

Let the sum of the series be k,


Thus,
$k = 1 + \dfrac{3}{2} + \dfrac{5}{4} + \dfrac{7}{8} + ..... + \dfrac{2n-1}{2^{n - 1}} \: \: \: \: \: \: \: \it{...(i)}$


Divide by ( 1 / 2 ) on both sides, [ as 1 / 2 is the common ratio of the geometric progression ].

Now,

$\dfrac{k}{2} = \dfrac{1}{2} \bigg \{1 + \dfrac{3}{2} + \dfrac{5}{4} + \dfrac{7}{8} + ..... + \dfrac{2n-3}{2^{n - 2} }+ \dfrac{2n-1}{2^{n - 1}} \bigg\} \\ \\ \\ \dfrac{k}{2} = \dfrac{1}{2} + \dfrac{3}{4} + \dfrac{5}{8} + \dfrac{7}{16} + ..... + \dfrac{2n-1}{2^{n - 1}} + \dfrac{2n-1}{2^{n}}\: \: \: \: \: \: \: \it{...(ii)}$



Now, subtracting ( ii ) from ( i ),

$k - \dfrac{k}{2} = 1 + \frac{3}{2} - \frac{1}{2} + \frac{5}{4} - \frac{3}{4} + \frac{7}{8} - \frac{5}{8} + .....\: + \frac{2n-1}{2^{n - 1}} - \frac{2n-3}{2^{n - 1}} - \frac{2n-1}{2^{n}} \\ \\ \\ k \bigg( \frac{1}{2} \bigg) = 1 + \bigg \{ \frac{2}{2} + \frac{2}{4} + \frac{2}{8} ..... + \frac{2}{2^{n - 1}} \bigg \} - \frac{2n-1}{2^{n}} \\ \\ \\ k \bigg( \frac{1}{2} \bigg) = 1 + \bigg \{1 + \frac{1}{2} + \frac{1}{4} + ..... + \frac{1}{2^{n - 2}} \bigg \} - \frac{2n-1}{2^{n}}$


Now, there is an another GP, 1 , 1 / 2, 1 / 4, 2 / 8.... 1 / ( 2^{ n - 2 } ).

Sum of GP = [ a( 1 - r^{ n - 1 } ) ] / [ 1 - r ]

Sum of new GP = [ 1( 1 - { 1 / 2 }^{ n - 1 } ] / [ 1 - 1 / 2 ]


Sum of new GP = 2[ 1 - ( 1 / 2 )^{ n - 1 } ]


Sum of new GP = 2 - [ 2 / 2^{ n - 1 } ]



Thus,


$k \times \dfrac{1}{2} = 1 + 2 - \dfrac{2}{{2}^{n - 1} } - \dfrac{2n-1}{2^{n}} \\ \\ \\ k \times \dfrac{1}{2} = 3 - \frac{{ {2}^{2} } }{2 {}^{n} } - \dfrac{2n - 1}{2 {}^{n} } \\ \\ \\ k \times \dfrac{1}{2} = \dfrac{3.2 {}^{n} - 4 - 2n + 1}{2 {}^{n} } \\ \\ \\ k = 2 \times \frac{3.2 {}^{n} - 2n - 3}{2 {}^{n} } \\ \\ \\ k = \dfrac{6.2 {}^{n} - 4n - 6}{2 {}^{n} } \: \: or \: \: 6 - \dfrac{4n + 6}{2 {}^{n} } \: \: or \: \: 6 - \dfrac{2n + 6}{2 {}^{n - 1} }$



Therefore the sum to $\mathsf{n}$ terms of the series $\mathsf{ 1 \: + \: \dfrac{3}{2} \: + \: \dfrac{5}{4} \: + \: \dfrac{7}{8}+.....} \:\:is\:\: \dfrac{6.2 {}^{n} - 4n - 6}{2 {}^{n} } \: \: or \: \: 6 - \dfrac{4n + 6}{2 {}^{n} } \: \: or \: \: 6 - \dfrac{2n + 6}{2 {}^{n - 1} }$