Question

1+ 2+ 3........99+100 =

Answer 1

Hey mate ^_^

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Solution:
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It is an AP where a = 1, d = 1 and n = 100.

The sum is S100 =(n/2)[2a + (n-1)d]

= (100/2)[2*1 + (100–1)*1]

= 50 [2+99]

= 50*101

= 5050

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Answer: 5050
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#Be Brainly❤️

Answer 2

Sum of first 100 terms(S)=1+2+3+…+98+99+100—(eq. i)

Similarly,

S=100+99+98+…+3+2+1—(eq. ii)

Adding (eq. i) and (eq. ii), he got,

S=1+2+3+…+98+99+100

+

S=100+99+98+…+3+2+1

(S+S)=(1+100)+(2+99)+(3+98)+…+(98+3)+(99+2)+(100+1)

∴2(S)=101+101+101+…+101+101+101(ie 101x100)

2(S)=101x100

∴S=101x100/2

∴S=101x50

∴S=5050.