100(1+100)

Example : 1+2+3+…+100= -----------
2

Simplify: (x+1) + 2(x+1)+ …+1000(x+1)

If you in google type:

Sum of numbers 1 to 100

By the way:

1 + 2 + 3 + 4 + . . . . + n = ( 1 + n )*( n / 2 )

1 + 2 + 3 +…100

n = 100

1 + 2 + 3 +…100 =( 1 + 100 ) * ( 100 / 2 ) = 101 * 50 = 5050

( x + 1 ) + 2 ( x + 1) + …+ 1000 ( x + 1 ) = ( x + 1 ) ( 1 + 2 +...1000 )

n = 1000

1 + 2 +....1000 = ( 1 + 1000 ) ( 1000 / 2 ) = 1001 * 500 = 500500

( x + 1 ) + 2 ( x + 1) + …+ 1000 ( x + 1 ) =

( x + 1 ) ( 1 + 2 +...1000 ) =

( x + 1 ) * 500500 = 505050 ( x + 1 )

( x + 1 ) * 500500 = 500500 ( x + 1 )

To simplify the expression (x+1) + 2(x+1) + … + 1000(x+1), we can use the distributive property of multiplication over addition.

Step 1: Distribute the coefficients to each term within the parentheses:
(x+1) + 2(x+1) + … + 1000(x+1) simplifies to
x + 1 + 2x + 2 + … + 1000x + 1000

Step 2: Combine like terms by adding the coefficients:
The x terms can be combined together:
(x + 2x + … + 1000x) + (1 + 2 + … + 1000)

To find the sum of the x terms, we need to sum the arithmetic series 1 + 2 + … + 1000.

Step 3: Find the sum of the arithmetic series:
The sum of an arithmetic series can be found using the formula:
S = (n/2)(a + l)
where S is the sum, n is the number of terms, a is the first term, and l is the last term.

In this case:
n = 1000 (since there are 1000 terms)
a = 1 (first term)
l = 1000 (last term)

Substituting these values into the formula, we have:
S = (1000/2)(1 + 1000) = 500 * 1001 = 500500

Therefore, the sum of the arithmetic series 1 + 2 + … + 1000 is 500500.

Step 4: Simplify the expression by substituting the sum of the arithmetic series:
x + 1 + 2x + 2 + … + 1000x + 1000 becomes:
(x + 2x + … + 1000x) + 500500

The terms with x can be combined together:
(x + 2x + … + 1000x) = x(1 + 2 + … + 1000)
Now we can substitute the sum of the arithmetic series:
x(1 + 2 + … + 1000) = x * 500500 = 500500x

Finally, the simplified expression is:
500500x + 500500