Important Summation Notations of Natural Number, Odd Number and Even Number Series

Dear Students, We are sharing some summations results of numbers which will help you in solving problems of algebra and trigonometry. Mainly, we focus on sum of number series, sum of square of number series and sum of cube of number series.

Shortcut / Formula 01: Sum of n Natural Numbers:

1+2+3+4+5+.....................................+n= n(n+1)/2

Example 01: 1+2+3+4+5+6+7+8 =?

Solution: Here n=8

so, 1+2+3+4+5+6+7+8  = 8(8+1)/2 = (8*9)/2 =72/2 = 36

Example 02: 1+2+3+...............+20 =?

Solution: Here n=20

so, 1+2+3+...............+20 = 20(20+1)/2 = (20*21)/2 =420/2 = 240

Shortcut / Formula 02: Sum of n Odd Numbers:

1+3+5+.....................................+n= n(n+1)

Example 01: 1+3+5+7+9+11=?

Solution: Here n=6

so, 1+3+5+7+9+11 = 6(6+1) = (6*7) =42

Example 02: 1+3+5+7+...............+21 =?

Solution: Here n=11

so, 1+3+5+7+...............+21 = 11(11+1) = (11*12) = 132

 

Shortcut / Formula 03: Sum of n Even Numbers:

2+4+6+.....................................+n= n²

Example 01: 2+4+6+8+10+12+14 =?

Solution: Here n=7

so, 2+4+6+8+10+12+14  =  7² = 49

Example 02: 2+4+6+..................+24 = ?

Solution: Here n=12

so, 2+4+6+..................+24  = 12² = 144

Shortcut / Formula 04: Sum of Square of first n Natural Numbers:

1²+2²+3²+4²+5²+.....................................+n²= [n(n+1)(2n+1)]/6

Example 01: 1²+2²+3²+4²+5²+6²+7²+8² =?

Solution: Here n=8

so, 1²+2²+3²+4²+5²+6²+7²+8²  = [8(8+1)(2*8+1)]/6

= [(8*9)(16+1)]/6 =(72*17)/6 =204

Example 02: 1²+2²+3²+...............+20² =?

Solution: Here n=20

so, 1²+2²+3²+...............+20² = [20(20+1)(20*2+1)]/6

= [(20*21)(40+1)]/6 =(420*41)/6 = 2870

Shortcut / Formula 05: Sum of square of first n Odd Numbers:

1²+3²+5²+.....................................+n²= n(4n²-1)/3

Example 01: 1²+3²+5²+7²+9²+11²=?

Solution: Here n=6

so, 1²+3²+5²+7²+9²+11²= 6(4*6²-1)/3

= 6(4*36-1)/3 = 2*(144-1)=2*143=286

Example 02: 1²+3²+5²+7²+...............+21² =?

Solution: Here n=11

so, 1²+3²+5²+7²+...............+21²= 11(4*11²-1)/3

= 6(4*121-1)/3 = 2*(484-1)=2*483=966

Shortcut / Formula 06: Sum of square of first n Even Numbers:

2²+4²+6²+.....................................+n²= [2n(n+1)(2n+1)]/3

Example 01: 2²+4²+6²+8²+10²+12²+14² =?

Solution: Here n=7

so, 2²+4²+6²+8²+10²+12²+14²  =  [2*7(7+1)(2*7+1)]/3

=[14*8(14+1)]/3= [112*15]/3 =112*5 = 560

Example 02: 2²+4²+6²+..................+24² = ?

Solution: Here n=12

so, 2²+4²+6²+..................+24²

=  [2*12(12+1)(2*12+1)]/3

=[24*13(24+1)]/3= [14*13*25]/3 =4550/3 =1516.67

Shortcut / Formula 07: Sum of cube of first n Natural Numbers:

1³+2³+3³+4³+5³+.....................................+n³= [n(n+1)/2]²

Example 01: 1³+2³+3³+4³+5³+6³+7³+8³ =?

Solution: Here n=8

so, 1³+2³+3³+4³+5³+6³+7³+8³  =[ 8(8+1)/2]²

= [(8*9)/2]² =[72/2]² = 36² =1296

Example 02: 1³+2³+3³+...............+20³ =?

Solution: Here n=20

so, 1³+2³+3³+...............+20³ = [20(20+1)/2]²

= [(20*21)/2]² =[420/2]²= [240]²=57600

Shortcut / Formula 08: Sum of cube of first n Odd Numbers:

1³+3³+5³+.....................................+n³= 2n²(n+1)²

Example 01: 1³+3³+5³+73+9³+11³=?

Solution: Here n=6

so, 1³+3³+5³+73+9³+11³ = 2*6²(6+1)²

= 2*36(7)² =72*49 =3528

Example 02: 1³+3³+5³+7³+...............+21³ =?

Solution: Here n=11

so, 1³+3³+5³+7³+...............+21³ = 2*11²(11+1)²

=2*121(12)²=242*144 =34848

Shortcut / Formula 09: Sum of cube of first n Even Numbers:

2³+4³+6³+.....................................+n³= n² (2n²-1)

Example 01: 2³+4³+6³+8³+10³+12³+14³ =?

Solution: Here n=7

so, 2³+4³+6³+8³+10³+12³+14³=  (2*7²-1)7² 

= 49*(2*49-1)=49(98-1)=49*97=4753

Example 02: 2³+4³+6³+..................+24³ = ?

Solution: Here n=12

so, 2³+4³+6³+..................+24³  = (2*12²-1)12² 

= 144(2*144-1)=144(288-1)=144*287=41328

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