$$ \sum_{r=1}^{11} r5^r = \frac{43\times5^{a}+ 5}{b}$$
Find (a+b)
This question was asked in an exam. I got the answer 28. However the answer given was 15.
Here is my attempt:
Let S = $ \sum_{r=1}^{11} r5^r $
So, S = $5^1 + 2\cdot5^2 + 3\cdot5^3 + 4\cdot5^4... + 11\cdot5^{11}$ - (i)
$5S$ = $\hspace{1.5cm} 5^2 + 2\cdot5^3 + 3\cdot5^4 + ... + 10\cdot5^{11}+ 11\cdot5^{12}$ - (ii)
Subtracting (ii) from (i),
$-4S$ = $5^1 + 5^2 + 5^3 + ... + 5^{11} - 11\cdot5^{12}$
$-4S$ = $\frac{5(5^{11} - 1)}{4} - 11\cdot5^{12}$ - (Applied Formula for Sum of GP)
$4S$ = $11\cdot5^{12} - \frac{5(5^{11}-1)}{4}$
$S$ = $\frac{44\cdot5^{12}-5(5^{11}-1)}{16}$ $S$ = $\frac{44\cdot5^{12}-5^{12}+5}{16}$ So, $S$ = $\frac{43\cdot5^{12}+5}{16}$ Hence, a+b = 12+16 = 28.
I cannot point out where am I wrong. Can anyone pointout where have I gone wrong?
Your method is fine, here is another:
For $x\neq 0,1$ let set $f(x)=\sum\limits_{r=0}^n x^r=\dfrac{x^{n+1}-1}{x-1}$
$f'(x)=\sum\limits_{r=1}^n rx^{r-1}=\dfrac 1x\sum\limits_{r=1}^n rx^r$
Therefore we find the general formula:
$$\sum\limits_{r=1}^n rx^r=xf'(x)=\frac{(nx-n-1)x^{n+1}+x}{(x-1)^2}$$
Applying for $n=11$ and $x=5$ we find $\sum\limits_{r=1}^{11} r5^r=\dfrac{(55-11-1)5^{12}+5}{(5-1)^2}=\dfrac{43\cdot5^{12}+5}{16}$