Powers of 1/a

$S=1+2(\frac{1}{5})+3(\frac{1}{5})^2+...$
$\frac{1}{5}S=1(\frac{1}{5})+2(\frac{1}{5})^2+3(\frac{1}{5})^3+...$
$\frac{4}{5}S=S-\frac{1}{5}S=1+\frac{1}{5}+(\frac{1}{5})^2+...$
$\frac{4}{25}S=\frac{1}{5}+(\frac{1}{5})^2+(\frac{1}{5})^3+...$
$\frac{16}{25}S=(\frac{4}{5}-\frac{4}{25})S=1$
$S=\frac{25}{16}$

Let $\frac {1} {5} = x$

Now the series becomes $S = 1 + 2x + 3x^2 + 4x^3 + \ldots eqn(i)$

Looking at the series, we can say that, it is of the form $nx^{n-1}$

We need to convert it to the form $nx^n$ , so we multiply both sides with x.

What now we get is $Sx = x + 2x^2 + 3x^3 + 4x^4 + \ldots eqn(ii)$

Subtracting eqn(ii) from eqn(i), we get $S - Sx = (1 + 2x + 3x^2 + \ldots) - (x + 2x^2 + 3x^3 + \ldots)$

$\Rightarrow S - Sx = 1 + x + x^2 + x^3 + \ldots$

Now taking, x common we get $S - Sx = 1 + x(1 + x + x^2 + x^3 + \ldots)$ , which is again the same series.

Now, by substitution $S - Sx = 1 + x(S - Sx)$

$S(1 - x) = 1 + x(S - Sx)$

$S(1 - x) - x(S - Sx) = 1$

$S(1 - x)^2 = 1$

Substituting the values of $x = \frac {1} {5}$

we get $S = \frac {25} {16} = \frac {a} {b}$

Hence, a+b = 41

We observe,

                  1 +  1/5 +(1/5)^2 + (1/5)^3+(1/5)^4...                   -(1)

                       +   1/5 +(1/5)^2+(1/5)^3+(1/5)^4+.....        -(2)

                    +            (1/5)^2+(1/5)^3+(1/5)^4+......     -(3)

                    +                           (1/5)^3+(1/5)^4+.....   -(4)

                     +                                     (1/5)^4+....    -(5)
                           ………………………………………………………………..........................
                          ………………………………………………………………..........................
                          ………………………………………………........................................          

                         =  S = 1 + 2 (1/5) + 3 (1/5)^2+4(1/5)^3+5(1/5)^4+......   

        (1)=>1/(1-1/5)=5/4
        (2) => (1/5)/1-1/5=1/4
        (3)=> (1/5)^2/{1-1/5}=1/20
        (4)=> (1/5)^3/1-1/5=1/100
        (5)=> (1/5)^4/1-1/5=1/500
         ...........................................................................

Now,

             (1) +(2)+(3)+ (4)+(5) +………………….     = S 

                                                   = (5/4)/1-1/5
                                                                                                                                             =  (5/4)/(4/5)

                                                            =25/16=a/b

Therefore, a + b = 41

From the given equation:

$S =1 + 2(\frac{1}{5}) + 3 (\frac{1}{5})^2 + 4 (\frac{1}{5})^3 + \ldots$

Multiply by $\frac{1}{5}$ yields:

$\frac{1}{5}S = (\frac{1}{5}) + 2 (\frac{1}{5})^2 + 3 (\frac{1}{5})^3 + \ldots$

Subtstract this new equation from the former equation term by term to obtain:

$\frac{4}{5}S = 1 + (\frac{1}{5}) + (\frac{1}{5})^2 + (\frac{1}{5})^3 + \ldots$

The right hand side is now a geometric series, which converges to $\frac{1}{1-\frac{1}{5}}=\frac{5}{4}$ .
Thus $\frac{4}{5}S=\frac{5}{4}$ . So $S = \frac{25}{16}$ .
Therefore $a+b = 25+16=41$ .

We notice that $S$ can be dissected as follows:

$S=S_0+S_1+S_2\cdots$ where

$S_0=1+1/5+(1/5)^2\cdots$ ,

$S_1=(1/5)+(1/5)(1/5)+(1/5)(1/5)^2\cdots$ ,

$S_2=(1/5)^2+(1/5)^2(1/5)\cdots$ ,

$\cdots$

and so on.

By using the formula to calculate the sum to infinity, $S_\infty = a/(1-r)$ , we have

$S_n=(1/5)^n/(4/5)$

therefore,

$S=(5/4)/(1-1/5)$

$=(5/4)(5/4)$

$=25/16$

$\therefore a+b=25+16=41$

As we know that the sum of an infinite GP can be given as: $\frac{a}{1-r} = a + ar + ar^2 + ar^3 + ..........$ where $r<1$ .

Differentiating this with respect to r we get $\frac{a}{(1-r)^2} = a + 2ar + 3ar^2 + 4ar^3 +............$

Putting $a = 1 , r = \frac{1}{5}$ we get the required series. The sum of the series is $S = \frac{1}{(1- \frac{1}{5})^2}$ $= \frac{25}{16}$

$a+b = 41$

Note-A.G.P== Arithmetico-geometric progression. The given series is a standard A.G.P, with the AP being 1,2,3...... and G.P being 1,1/5,(1/5)^2....... A standard approach for such problems is to write down the sum as a variable, given as "S" in the problem. S=1+2(1/5)+3(1/5)^2......equation (a) Now multiply S by the common ratio of the Geometric progression involved.In this case it is 1/5. So, S/5=1/5 + 2(1/5)^2 + 3(1/5)^3......equation (b). Now subtract equation (b) from equation (a).This is the "smart" step. Write the "1" in equation (a) as such and subtract the terms that have 1/5 raised to the same power.Giving 4S/5=1+ (1/5) + (1/5)^2 + (1/5)^3....... Now the Right hand side is a simple geometric progression whose sum can be evaluated using the standard G.P formula.That is, (first term)/1-(common ratio). It comes out to be 5/4 in this case. So 4S/5=5/4. S=25/16. Since 25 and 16 are coprime, the required sum=25+16=41. Hence the answer.

$\begin{array}{rllllll} S &= &1 &+ 2\left(\frac{1}{5}\right)& + 3\left(\frac{1}{5}\right)^2 &+ 4\left(\frac{1}{5}\right)^3 & \ldots \\ -\left(\frac{1}{5}\right)S &= & & -\left(\frac{1}{5}\right) &-2\left(\frac{1}{5}\right)^2 &-3\left(\frac{1}{5}\right)^3 & \ldots \\ \hline \left(\frac{4}{5}\right)S &=& 1 & + \left(\frac{1}{5}\right) & + \left(\frac{1}{5}\right)^2 & + \left(\frac{1}{5}\right)^3 & \ldots \\ \end{array}$

The last row is obtained by summing up the entries in each column. The right hand side is recognized as an infinite geometric series, so it equals $\frac{1}{1 - \frac{1}{5}} = \frac{5}{4}$ . Therefore $S = \frac{5}{4} \cdot \frac{5}{4} = \frac{25}{16}$ . Hence $a + b = 25 + 16 = 41$ .

Let $x = \frac{1}{5}$ . The series computes to:

$\Sigma_{n=1}^{\infty} n(1/5)^{n-1} = \frac{d}{dx}(\Sigma_{n=1}^{\infty} x^n) = \frac{d}{dx}(\frac{x}{1-x}) = \frac{1}{(1-x)^2} = \frac{1}{(1-1/5)^2} = \frac{25}{16}.$

Thus, $a = 25, b = 16 \Rightarrow a + b = \boxed{41}.$

Using Arithmetic-Geometric Progression , we have $a=1, r=\frac{1}{5}, \text{and}~d=1$

$S=\dfrac{a}{1-r}+\dfrac{d\times{r}}{\left(1-r\right)^2}=\dfrac{1}{1-\frac{1}{5}}+\dfrac{1\times{\frac{1}{5}}}{\left(1-\frac{1}{5}\right)^2}=\dfrac{5}{4}+\dfrac{5}{16}=\dfrac{25}{16}$ .

Then the required answer is $~~\boxed{41}$

41