Hmm...V

$\begin{array}{c}~100+101+\ldots+199&=\sum_{i=100}^{199}i\\~\\~\\&=\sum_{i=0}^{199}i-\sum_{i=0}^{99}i\\~\\~\\&=\dfrac{199(200)}{2}-\dfrac{99(100)}{2}\\~\\~\\&=\boxed{14950} \end{array}$

The Sum of Natural number up to N is given by {(N) * (N+1)}/2

Sum of Natural number up to 199 is i.e. S1 = (199 * 200)/2 = 19900

Sum of Natural number up to 99 is i.e. S2 = (99 * 100)/2 = 4950

The Required Sum, i.e. S = S1- S2 = 14950

arithmetic series apply equation, S = n/2 {2a + (n-1)d} Here, n = 100; a = 100; d =1

This is just an Arithmetic Series

$n = 100, t_{1} = 100, t_{100} = 199$

$S_{n} = \frac{n}{2}(t_{1} + t_{n})$

$S_{100} = \frac{100}{2}(100 + 199)$

$S_{100} = 50 (299)$

$S_{100} = \boxed{14950}$

100,101,102.....................,199
+
199,198,197.....................,100
299+299+299....................+299=(299*100)/2=14950

$s=\dfrac{n}{2}[2a_1+(n-1)d]$

$s=\dfrac{100}{2}[2(100)+99]=14950$

Let $n =$ Number of terms , $a =$ First term and $d =$ Common difference

The formula for working this out is

$na + \frac{dn(n - 1)}{2}$

This is proved here .

$n = 100$

$a = 100$

$d = 1$

So let's input the values

$(100)(100) + \frac{(1)(100)((100)-1)}{2} = 10,000 + \frac{9900}{2}$

$10,000 + 4950 = 14,950$

first 3 digits number: 100 to 199

100+(100+1)+(100+2)+...+(100+99)

=100*100+(1+2+3+...+99)

match it like this:

=10000+((99+1)+(98+2)+...+50)

since only one option that end with 50, the answer is 14950

100+199=101+198=102+197.......154+155=299

So if there are 50 pairs, 299*50=14950

First 100 three digit numbers: 100,101,102,...,199. Arithmetic series: a=100, d=1, n=100. Use formula: n/2[2a+(n-1)d]=50(299)=14950.

just using summation

a.p formula n/2[2a+(n-1)d] where n=100 a=100 and d=1

by putting the value we get 14950

the formula is Sn=n/2(2a+(n-1)d) where a=100,n=100,d=1

Gauss solved a similar problem when in grade school hundreds of years ago. You add 100 plua 199=299, and then since you repeat that through all the rest of the numbers, 101 plus 198=299, you have the sum 299 taken 50 times=29900 divided by 2=14950

1. Knowing 100-199 is the first one hundred three digit numbers means 149 is the middle number.

2. Multiply this by 100 (as there are one hundred numbers) to get 14900 as a rough average.

3. then adding the number of ones that are left for the larger 50 numbers bigger than the average of 150-199. (each larger number has its extra ones that are greater than the average numbers "distributed" to the lower numbers than 149 until you get to 199 which is left without a lower number to give its ones to. This has 50 ones left over.

4. This gives you 14900+50=14950

I honestly guessed based on reasoning