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