Question

If the sum of $1+\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+\cdots n$ terms $=S=k-\frac{mn+3}{m^{n-1}}$, then find $k-m$ ?

Answer 1

Let $S=1+\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+\cdots n$ terms

$S=1+\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+\cdots$

$S=\frac{2\times 1-1}{2^{1-1}}+\frac{2\times 2-1}{2^{2-1}}+\frac{2\times 3-1}{2^{3-1}}+\cdots$

So we can see that $a_n=\frac{2n-1}{2^{n-1}}$

$a_n=\frac{4n-2}{2^n}$

Now we divide this question into two parts

$b_n=\frac{n}{2^n}$

$S_1=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots \frac{n}{2^n}$

$\frac{S_1}{2}=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\cdots \frac{n}{2^{n+1}}$

$\frac{S_1}{2}=1-{\frac{1}{2}}^n-\frac{n}{2^{n+1}}$
$S_1=2-\frac{1}{2^{n-1}}-\frac{n}{2^n}$
$c_n=\frac{1}{2^n}$
$S_2=1-{\left(\frac{1}{2}\right)}^n$
$S=4S_1-2S_2$
$S=4(2-\frac{1}{2^{n-1}}-\frac{n}{2^n})-2(1-{\frac{1}{2}}^n)$
$S=8-\frac{(4n+8)}{2^n}-2+\frac{2}{2^n}$
$S=6-\frac{2n+3}{2^{n-1}}$
$k=6$ and $m=2$