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Find the value of Sn=\( \frac{1}{2} + \frac{3}{2^2} + \frac{5}{2^3} + ... + \frac{2n-3}{2^{n-1}} + \frac{2n-1}{2^n}\)

 May 18, 2022
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Find the value of
\(S_n=\dfrac{1}{2} + \dfrac{3}{2^2} + \dfrac{5}{2^3} + ... + \dfrac{2n-3}{2^{n-1}} + \dfrac{2n-1}{2^n}\)

 

\(\text{Let $x=\dfrac12$}\)

 

\(\begin{array}{|rcll|} \hline S_n&=&x + &3x^2 + 5x^3+ 7x^4 + ... + (2n-3)x^{n-1} + (2n-1)x^n \\ xS_n&=& &1x^2 + 3x^3+ 5x^4 + ... + (2n-5)x^{n-1} + (2n-3)x^n+(2n-1)x^{n+1} \\ \hline x-xS_n &=& x + &2x^2 + 2x^3+ 2x^4 + ... + 2x^{n-1} + 2x^n -(2n-1)x^{n+1} \\ S_n(1-x) &=& x + &2*(x^2 + x^3+ x^4 + ... + x^{n-1} + x^n) -(2n-1)x^{n+1} \\ &&& \boxed{x^2 + x^3+ x^4 + ... + x^{n-1} + x^n = \dfrac{x^2-x^{n+1} } {1-x} } \\ S_n(1-x) &=& x + &2*\dfrac{x^2-x^{n+1} } {1-x} -(2n-1)x^{n+1} \quad | \quad 1-x=1-\dfrac12=\dfrac12 \\ S_n*\dfrac12 &=& x + &4(x^2-x^{n+1}) -(2n-1)x^{n+1} \quad | \quad * 2 \\ S_n &=& 2x + &8(x^2-x^{n+1}) -2(2n-1)x^{n+1} \\ S_n &=& 2x + &8x^2-8x^{n+1} -2(2n-1)x^{n+1} \\ S_n &=& 2x + &8x^2-2x^{n+1}\Big(4 +(2n-1)\Big) \\ S_n &=& 2x + &8x^2-2x^{n+1}(2n+3) \\ S_n &=& 2x \Big(& 1 + 4x-x^n(2n+3)\Big) \quad | \quad x=\dfrac12 \\ S_n &=& &1 + 2-\dfrac{2n+3}{2^n} \\ \mathbf{S_n} &=& \mathbf{3-}&\mathbf{\dfrac{2n+3}{2^n}} \\ \hline \end{array} \)

 

laugh

 May 19, 2022

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