Introduction for summation math:
Summation math is assumed to be the sequence form of addition. Summation is the process of combine a sequence of numbers by means of addition. The summation can be written as n1 + n2 + n3 + n4 + n5 + ……… nn. Summation math is a continuous function for one variable in closed interval. Summation of an infinite series of value is not constantly possible. In this topic we will discuss the some examples problems.
Summation Math Rules:
Summation of infinite numbers is called summation,
`sum_(n=1)^oo` = 1 + 2 + 3 + 4 + 5 + 6 + ...... + `oo`
Summation Rules:
Summation of a number
`sum_(n=1)^oo` a = a + a + a + ...... a (infinite times) = a * (`oo` )
Summation of square of infinite number.
`sum_(n =1)^oo` i2 = 12 + 22 + 32 + 42 + 52 + ….. n2 = `(n (n + 1) (2n + 1))/(6)`
Summation of cube of infinite number.
`sum_(n=1)^oo` i3 = 13 + 23 + 33 + 43 + 53 + ….. n3 = `(n^2 (n + 1)^2/4) `
`sum_(i=1)^n` si = s1 + s2 + s3 + … + sn
(n times) = sn, where s is constant.
`sum_(i=1)^n` i = 1 + 2 + 3 + … + n = `(n (n +1)) /(2) `
Example for Summation Math:
Example 1: Determine the value of the summation math `sum_(n=1)^4` ` (3 + 4n)`
Solution:
`sum_(n=1)^oo` `= 1 + 2 + 3 + 4 + 5 + 6 + ...... + n`
`sum_(n=1)^5` `(3 + 4n) = (3 + 4(1)) + (3 + 4(2)) + (3 + 4(3)) + (3 + 4(4)) `
`= (3 + 4) + (3 + 8) + (3 + 12) + (3 + 16) `
`= 7 + 11 + 15 + 19 = 52`
Answer: 52
Example 2: Determine the value of the summation math `sum_(n=1)^6` `(n + 1)^2`
Solution:
Here, `(n + 1)^2`
`(n + 1)^2 = n^2 + 2n +1`
so, we wirte the summation as `sum_(n=1)^6` ` (n^2 + 2n+ 1)`
the summation, we get
`sum_(n=1)^6` `n^2` + `sum_(n=1)^6` `2n` + `sum_(n=1)^6`
`= {12 + 22 + 32 + 42 + 52 + 6 ^2} + {(2 * (1)) + (2 * (2)) + (2 * (3)) + + (2 * (4)) + + (2 * (5)) + + (2 * (6))} + (1 * (6))`
= `((6 (6 + 1) ((2 * (6)) + 1))/(6)) + 42 + 6`
`=91 + 42 + 6`
Answer: 139
Summation math is assumed to be the sequence form of addition. Summation is the process of combine a sequence of numbers by means of addition. The summation can be written as n1 + n2 + n3 + n4 + n5 + ……… nn. Summation math is a continuous function for one variable in closed interval. Summation of an infinite series of value is not constantly possible. In this topic we will discuss the some examples problems.
Summation Math Rules:
Summation of infinite numbers is called summation,
`sum_(n=1)^oo` = 1 + 2 + 3 + 4 + 5 + 6 + ...... + `oo`
Summation Rules:
Summation of a number
`sum_(n=1)^oo` a = a + a + a + ...... a (infinite times) = a * (`oo` )
Summation of square of infinite number.
`sum_(n =1)^oo` i2 = 12 + 22 + 32 + 42 + 52 + ….. n2 = `(n (n + 1) (2n + 1))/(6)`
Summation of cube of infinite number.
`sum_(n=1)^oo` i3 = 13 + 23 + 33 + 43 + 53 + ….. n3 = `(n^2 (n + 1)^2/4) `
`sum_(i=1)^n` si = s1 + s2 + s3 + … + sn
(n times) = sn, where s is constant.
`sum_(i=1)^n` i = 1 + 2 + 3 + … + n = `(n (n +1)) /(2) `
Example for Summation Math:
Example 1: Determine the value of the summation math `sum_(n=1)^4` ` (3 + 4n)`
Solution:
`sum_(n=1)^oo` `= 1 + 2 + 3 + 4 + 5 + 6 + ...... + n`
`sum_(n=1)^5` `(3 + 4n) = (3 + 4(1)) + (3 + 4(2)) + (3 + 4(3)) + (3 + 4(4)) `
`= (3 + 4) + (3 + 8) + (3 + 12) + (3 + 16) `
`= 7 + 11 + 15 + 19 = 52`
Answer: 52
Example 2: Determine the value of the summation math `sum_(n=1)^6` `(n + 1)^2`
Solution:
Here, `(n + 1)^2`
`(n + 1)^2 = n^2 + 2n +1`
so, we wirte the summation as `sum_(n=1)^6` ` (n^2 + 2n+ 1)`
the summation, we get
`sum_(n=1)^6` `n^2` + `sum_(n=1)^6` `2n` + `sum_(n=1)^6`
`= {12 + 22 + 32 + 42 + 52 + 6 ^2} + {(2 * (1)) + (2 * (2)) + (2 * (3)) + + (2 * (4)) + + (2 * (5)) + + (2 * (6))} + (1 * (6))`
= `((6 (6 + 1) ((2 * (6)) + 1))/(6)) + 42 + 6`
`=91 + 42 + 6`
Answer: 139
No comments:
Post a Comment