Using the Gauss method to add 1+2+3+...+98+99+100= 50(1+100)=5050, Find the sum of 1x+2+3x+4+5x+...+98+99x+100.

Using the Gauss method to add 1+2+3+...+98+99+100= 50(1+100)=5050,

Find the sum of  1x+2+3x+4+5x+...+98+99x+100.

Can you please show how you got the answer as well, thank you.

$$\small{\text{
$
1x+2+3x+4+5x+ \dots +98+99x+100 =
$
}}\\
\small{\text{
$
\underbrace{
[\ 1x+3x+5x+7x+ \dots + 97x+99x \ ]
}_{\text{Part}\ 1}
+
\underbrace{
[ 2+4+6+8+\dots +98+100\ ]
}_{\text{Part}\ 2}
$
}}$$

Part 1:

$$\small{\text{
$[\ 1x+3x+5x+7x+ \dots + 97x+99x \ ]
=
2x[\ \underbrace{ 1+2+3+4+\dots + 49+ 50}_{=(1+50)*\frac{50}{2}}\ ] -50x $
}}\\
\small{\text{
$=(1+50) *50x-50x = 2550x-50x = 2500x
$
}}$$

Part 2:

$$\small{\text{
$
[ 2+4+6+8+\dots +98+100\ ] = 2*[\ \underbrace{ 1+2+3+4+\dots + 49+ 50}_{=(1+50)*\frac{50}{2}}\ ]
$
}}\\
\small{\text{
$=(1+50) *50 = 2550
$
}}$$

Part 1 + Part 2 = 2550 + 2500x

Let's work on the 'x" terms, first

Notice that

1x + 99x = 100x

And

3x + 97x = 100x

So, it appears that adding all the x terms pair-wise like this will produce 100x. And we have 25 of them...so 25 x 100x = 2500x

Likewise

2 + 100, 4 + 98, 6 + 96.... all produce 102......and we have 25 pairs of these. So 102 x 25 = 2550

So, the "series" sums to   2550 + 2500x