+0

# 101+102+...201=?

#1
+8

This problem is an altered form of an easier problem, which I will explain first before I'll move on to your problem.

Suppose we want to add the numbers 1 to 10.

$$\begin{array}{lcl} 1+1 = 2\\ 2+3 = 5\\ 5+4 = 9\\ 9+5 = 14\\ \mbox{etc.} \end{array}$$

Since this takes quite a long time, we can also try to find a more logical way to solve this.

Let's try to find pairs which make it easier for us to solve this summation.

$$\begin{array}{lcl} 0+10 = 10\\ 1+9 = 10\\ 2+8 = 10\\ 3+7 = 10\\ 4+6 = 10\\ \ \ \ \ \ \ 5 = 5\\ \_\_\_\_\_\_\_\_\_\_\_\_\_ \\ 1+...+10 = 55\\ \mbox{Hence the sum of all the numbers between 1 and 10 is 55}\\ \end{array}$$

$$\begin{array}{lcl} \mbox{Note that we could also write this as } 10\times \frac{10}{2}+\frac{10}{2}\\ \mbox{This again can be rewritten to} \frac{10 \times 10 + 10}{2}\\ \mbox{and } \frac{10 \times 10 + 10}{2} = \frac{10 \times (10+1)}{2} \end{array}$$

Try to see for yourself that this idea doesn't just hold for the sum of the numbers 1 to 10, but also for numbers 1 to n. where n is any number you want to choose. Therefore

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

Now find the sum of 1 to 201, the sum of 1 to 100 and then substract the second from the first :D

Then you can find the sum of 1 to 201.

Good luck!

Nov 11, 2014

#1
+8

This problem is an altered form of an easier problem, which I will explain first before I'll move on to your problem.

Suppose we want to add the numbers 1 to 10.

$$\begin{array}{lcl} 1+1 = 2\\ 2+3 = 5\\ 5+4 = 9\\ 9+5 = 14\\ \mbox{etc.} \end{array}$$

Since this takes quite a long time, we can also try to find a more logical way to solve this.

Let's try to find pairs which make it easier for us to solve this summation.

$$\begin{array}{lcl} 0+10 = 10\\ 1+9 = 10\\ 2+8 = 10\\ 3+7 = 10\\ 4+6 = 10\\ \ \ \ \ \ \ 5 = 5\\ \_\_\_\_\_\_\_\_\_\_\_\_\_ \\ 1+...+10 = 55\\ \mbox{Hence the sum of all the numbers between 1 and 10 is 55}\\ \end{array}$$

$$\begin{array}{lcl} \mbox{Note that we could also write this as } 10\times \frac{10}{2}+\frac{10}{2}\\ \mbox{This again can be rewritten to} \frac{10 \times 10 + 10}{2}\\ \mbox{and } \frac{10 \times 10 + 10}{2} = \frac{10 \times (10+1)}{2} \end{array}$$

Try to see for yourself that this idea doesn't just hold for the sum of the numbers 1 to 10, but also for numbers 1 to n. where n is any number you want to choose. Therefore

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

Now find the sum of 1 to 201, the sum of 1 to 100 and then substract the second from the first :D

Then you can find the sum of 1 to 201.

Good luck!

reinout-g Nov 11, 2014
#2
+5

$$Sn = \frac{(a1+an)*n}{2} S = the sum of the numbers a1 = first number an = last number n = quantity of numbers an = a1 + d(n-1) [d = 1 = the common difference of successive members) 201 = 101 - 1(n-1) n = 101 S101 = [(101+201)*101]/2 S101 = 15251$$

.
Nov 11, 2014