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5 mar 2014 · Mathematical Induction: Example 11) Prove that 1^3 + 2^3 +3^3 ... n^3 = n^2 {(n+1)^2} /4
HINT: You want that last expression to turn out to be (1 + 2 + … + k + (k + 1))2, so you want (k + 1)3 to be equal to the difference. (1 + 2 + … + k + (k + 1))2 − (1 + 2 + … + k)2. That’s a difference of two squares, so you can factor it as. (k + 1)(2(1 + 2 + … + k) + (k + 1)).
9 gru 2014 · The result now follows immediately by F(n) = (n(n + 1) / 2)2 ⇒ F(n) − F(n − 1) = n3. The theorem reduces the proof to a trivial mechanical verification of a polynomial equality, which requires no ingenuity whatsoever.
15 paź 2019 · $1^3+2^3+...+n^3=$ has a formula which $(1+2+3+...+n)^2 $ or $[n(n + 1)/2]^2$. We can verify with ınduction and I know how I can prove it. How we find what is the formula...
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16 kwi 2024 · Question2: Prove the following by using the principle of mathematical induction 13 + 23 + 33+ + n3 = ( ( +1)/2)^2 Let P (n) : 13 + 23 + 33 + 43 + ..+ n3 = ( ( +1)/2)^2 For n = 1, L.H.S = 13 = 1 R.H.S = (1(1 + 1)/2)^2= ((1 2)/2)^2= (1)2 = 1 Hence, L.H.S. = R.H.S P(n) is true for n = 1 Assume that P(k) is true 13 + 23 + 33 + 43 + ..+ k3 = ( ( + 1 ...
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