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$\begin{aligned} S & = \sum_{n=1}^{100} \frac {(-1)^n(5n-1)}{10} \\ & = \sum_{n=1}^{100} (-1)^n\left(\frac n2 - \frac 1{10}\right) \\ & = \left(-\frac 12 + 1 - \frac 32 + 2 - \frac 52 + 3 - \cdots - \frac {99}2 + 50\right) + \left(\frac 1{10} - \frac 1{10} + \frac 1{10} - \frac 1{10} + \frac 1{10} - \frac 1{10} + \cdots - \frac 1{10} \right) \\ & = - \frac 12 (1+3+5+\cdots + 99) + (1+2+3+\cdots + 50) + 0 \\ & = - \frac 12 \cdot \frac {50}2 (1+99) + \frac {50(50+1)}2 \\ & = - 1250 + 1275 = 25 \end{aligned}$

Therefore $\log_5 S = \log_5 25 = \boxed 2$ .

• We first look at the summation part. It gets really easy when we split these into even $(2n)$ and odd $(2n-1)$ and negative sign-ed

$\sum_{n=1}^{100} \dfrac{(-1)^n (5n-1)}{10} = \blue{\sum_{n=1}^{50} \dfrac{(5(2n)-1)}{10}} - \orange{\sum_{1}^{50} \dfrac{(5(2n-1)-1)}{10}} \quad \ldots (1)$

• Now, we evaluate both of these separately

$\blue{\sum_{n=1}^{50} \dfrac{(5(2n)-1)}{10}} =\dfrac{\frac{n}{2}(a+l)}{10} = \blue{\dfrac{50}{20}{(9 + 499)}}$

$\orange{\sum_{1}^{50} \dfrac{(5(2n-1)-1)}{10}} = \dfrac{\frac{n}{2}(a+l)}{10} = \orange{\dfrac{50}{20}{(4 + 494)}}$

• We substitute back into equation $(1)$

$\sum_{n=1}^{100} \dfrac{(-1)^n (5n-1)}{10} = \blue{\dfrac{50}{20}{(9 + 499)}} - \orange{\dfrac{50}{20}{(4 + 494)}} = \dfrac{5}{2}(10) = 25$

• $\log_{5}$ of the entire expression at last

$\log_{5} \bigg(\sum_{n=1}^{100} \dfrac{(-1)^n (5n-1)}{10} \bigg) = \log_{5} 25 = \boxed{2}$

For odd $n$ 's, ${ -1 }^{ n }$ would be -1 only, so the numerator would be just $-1 \times (5n-1)=-5n+1$ and for even $n$ 's, ${ -1 }^{ n }$ would be 1, so the numerator would be $1 \times (5n-1)=5n-1$ .

Also, there are 50 even numbers and 50 odd numbers in the first 100 natural numbers, which means for every even number there is an odd number. Suppose, there is an even number $2n$ so as per the function the output or the term would be $\dfrac{ 5(2n)-1 }{ 10 }$ and for an odd number before it, i.e $2n-1$ the term would be $\dfrac{ -5(2n)+1 }{ 10 }$ as we saw the denominator for an odd term. If we add both we obtain the sum as,

$\frac{ -5(2n)+1 }{ 10 }+\frac{ 5(2n)-1 }{ 10 }=\frac{ -5(2n-1)+1+5(2n)-1 }{ 10 }=\frac{ -5(-1) }{ 10 }=\frac{ 5 }{ 10 }=\frac{ 1 }{ 2 }$

So for each odd-even pair, the sum is $\frac{ 1 }{ 2 }$ , as there are 50 such pairs for all the first 100 natural numbers, we are adding $\dfrac{ 1 }{ 2 }$ 50 times, which is $\frac{ 1 }{ 2 } \times 50=25$ .

So, $\log _{ 5 }{ 25 }$ is 2 as ${ 5 }^{ 2 }=25$

Every 2 terms in the summation subtract to $\frac12$ . Thus the summation equals 50 terms of $\frac12$ , which is 25. $\log_5 25 = \boxed{2}$ .

$\begin{aligned} \left( \displaystyle\sum_{n=1}^{100} \cfrac{(-1)^n (5n-1)}{10} \right) &= \left( \displaystyle\sum_{n=1}^{50} \cfrac{(-1)^{2n-1} (5(2n-1)-1)}{10} + \cfrac{(-1)^{2n} (5(2n)-1)}{10} \right) \\ &= \left( \displaystyle\sum_{n=1}^{50} \cfrac{(-1) (10n-5-1)}{10} + \cfrac{ (10n-1)}{10} \right) \\ &= \left( \displaystyle\sum_{n=1}^{50} \cfrac{ (-10n+5+1) + (10n - 1)}{10} \right) \\ &= \left( \displaystyle\sum_{n=1}^{50} \cfrac{5}{10} \right) = 50 \cdot \cfrac{5}{10} = 25 \\ \log_5 \left( \displaystyle\sum_{n=1}^{100} \cfrac{(-1)^n (5n-1)}{10} \right) &= \log_5 (25) = \boxed{2} \end{aligned}$

Taking the summation part only we see

$n=1 \rightarrow \dfrac{(-1)^1(5*1 - 1)}{10} \rightarrow -0.4$

$n=2 \rightarrow \dfrac{(-1)^2(5*2 - 1)}{10} \rightarrow 0.9$

$n=3 \rightarrow \dfrac{(-1)^3(5*3 - 1)}{10} \rightarrow -1.4$

$n=4 \rightarrow \dfrac{(-1)^4(5*4 - 1)}{10} \rightarrow 1.9$

$........$

$n=99 \rightarrow \dfrac{(-1)^{99}(5*99 - 1)}{10} \rightarrow -49.4$

$n=100 \rightarrow \dfrac{(-1)^{100}(5*100 - 1)}{10} \rightarrow 49.9$

Since the values alternate between positive and negative, we'll make two arithmetic progressions, one where $n$ has been odd in terms and another with positive $n$ , and take two distinct sums.

$AP_1 \rightarrow -0.4, -1.4, ......, -49.4 \rightarrow S_{50}= \frac{50}{2}[-0.4-49.4]= -1245$

$AP_2 \rightarrow 0.9, 1.9, ......, 49.9 \rightarrow S_{50} = \frac{50}{2}[0.9+49.9]= 1270$

Now we can determine the whole sum= $1270+ (-1245) = 25$

Since $5^2 =25$ , $\text{log}_5 25 =\boxed{2}$