Prove 1 + 2 + 3 ... + n = n(n+1)/2 - Mathematical Induction - Theory

part 2 - Addition - Theory - Serial order wise - Mathematical Induction
part 3 - Addition - Theory - Serial order wise - Mathematical Induction

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Prove 1 + 2 + 3 + โ€ฆโ€ฆ. + n = (๐ง(๐ง+๐Ÿ))/๐Ÿ for n, n is a natural number Step 1: Let P(n) : (the given statement) Let P(n): 1 + 2 + 3 + โ€ฆโ€ฆ. + n = (n(n + 1))/2 Step 2: Prove for n = 1 For n = 1, L.H.S = 1 R.H.S = (๐‘›(๐‘› + 1))/2 = (1(1 + 1))/2 = (1 ร— 2)/2 = 1 Since, L.H.S. = R.H.S โˆด P(n) is true for n = 1 Step 3: Assume P(k) to be true and then prove P(k + 1) is true Assume that P(k) is true, P(k): 1 + 2 + 3 + โ€ฆโ€ฆ. + k = (๐‘˜(๐‘˜ + 1))/2 We will prove that P(k + 1) is true. P(k + 1): 1 + 2 + 3 +โ€ฆโ€ฆ. + (k + 1) = ((k + 1)( (k + 1) + 1))/2 P(k + 1): 1 + 2 + 3 +โ€ฆโ€ฆ.+ k + (k + 1) = ((๐ค + ๐Ÿ)(๐ค + ๐Ÿ))/๐Ÿ We have to prove P(k + 1) is true Solving LHS 1 + 2 + 3 +โ€ฆโ€ฆ.+ k + (k + 1) From (1): 1 + 2 + 3 + โ€ฆโ€ฆ. + k = (๐‘˜(๐‘˜ + 1))/2 = (๐’Œ(๐’Œ + ๐Ÿ))/๐Ÿ + (k + 1) = (๐‘˜(๐‘˜ + 1) + 2(๐‘˜ + 1))/2 = ((๐’Œ + ๐Ÿ)(๐’Œ + ๐Ÿ))/๐Ÿ = RHS โˆด P(k + 1) is true when P(k) is true Step 4: Write the following line Thus, By the principle of mathematical induction, P(n) is true for n, where n is a natural number

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