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Summation notation is a convenient way to write the sum of the terms of a finite sequences.

Because the Greek capital letter sigma $\sum$ is used to indicate the sum, the summation is also referred to as sigma notation.

As an example, consider the following: 

$$\sum^n_{r=1} (1+r)$$

This notation tells us to find the sum of (1+r) from r=1 to r=n

In the above, r is called the index of summation, 1 is the lower limit of the summation and n is the upper limit of the summation.

Example 1

Evaluate $$\sum^5_{n=2} n^2$$


According to the notation, the lower limit of summation is 2 and the upper limit is 5. So, we need to find the sum of the n² from n=2 to n=5. This is easily achieved by substituting n=2,3,4,5 into the notation.



Rules/properties of summation notation

Rule 1

$$\sum^n_{k=1} c=cn$$

where c is constant

Example 2

Solve $$\sum^5_{k=1}2$$


Here 2 is a constant, hence, we apply the rule above


Rule 2

$$\sum^n_{k=1}ck=c\sum^n_{k=1} k$$ where C is constant and k is variable

Example 3

Evaluate $$\sum^5_{k=2} 3k$$


$$\sum^5_{k=2} 3k=3\sum^6_{k=2} k$$

$$3\sum^6_{k=2} k=3(2+3+4+5)$$

$$3\sum^6_{k=2} k=3(14)=42$$

Rule 3

$$\sum^n_{k=1}(x+y)=\sum^n_{k=1}x +\sum^n_{k=1}y$$

Example 4

Evaluate $$\sum^3_{k=1}(k²+k)$$


According to rule 3,

$$\sum^3_{k=1}(k^2+k)=\sum^3_{k=1}k^2 +\sum^3_{k=1}k$$

$$\sum^3_{k=1}k^2 +\sum^3_{k=1}k=\left((1)^2+(2)^2+(3)^2\right)+(1+2+3)$$

$$\sum^3_{k=1}k^2 +\sum^3_{k=1}k=(1+4+9+1+2+3)=20$$

Rule 4

$$\sum^n_{k=1}(x-y)=\sum^n_{k=1}x -\sum^n_{k=1}y$$

Example 5

Evaluate $$\sum^3_{k=1}(k²-k)$$


According to rule 4,

$$\sum^3_{k=1}(k^2-k)=\sum^3_{k=1}k^2 -\sum^3_{k=1}k$$

$$\sum^3_{k=1}k^2 -\sum^3_{k=1}k=\left((1)^2+(2)^2+(3)^2\right)-(1+2+3))$$

$$\sum^3_{k=1}k^2 - \sum^3_{k=1}k=(1+4+9-(1+2+3))=8$$

$$\sum^n_{k=1}(x-y)=\sum^n_{k=1}x -\sum^n_{k=1}y$$

Having taken the example, we moved to more examples 

Example 6

Solve $$\sum^3_{k=1}(-1)^k k^2$$


$$\sum^3_{k=1}(-1)^k k^2=\left((-1)(1)^2)+((-1)^2(2)^2)+((-1)^3(3)^2\right)$$

$$\sum^3_{k=1}(-1)^k k^2=(-1)+(4)+(-9)=-6$$

Related post

Example 7

Solve $$\sum^5_{n=2}(3n-1)$$




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