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.
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/ 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
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/ 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
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/ 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
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/ 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
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/ 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
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/ 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
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/ 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
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/ 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
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/ 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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