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A.Maths (MI) 趕趕趕!!!!

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a. 1x5+2x6+3x7+....+n(n+4)=1/6n(n+1)(2n+13) 1+2+3+4....+n=1/2n(n+1) find 1X4+2X5+3x6+.....+n(n+3) and 1^2+2^2+......+n^2 (show working steps pls)

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a. 1x5+2x6+3x7+....+n(n +4)= n(n+1)(2n+13)/6 1+2+3+4....+n=n(n +1)/2 Let P(n) = 1x5+2x6+3x7+....+n(n +4)= n(n+1)(2n+13)/6 When n = 1 L.H.S = 1x5 = 5 R.H.S = n(n+1)(2n+13)/6 = 1(1+1)(2x1+13)/6 = 5 As L.H.S = R.H.S So P(1) is true Assume P(k) is true, i.e. 1x5+2x6+3x7+....+k(k+4)= k(k+1)(2k+13)/6 When n = k+1 L.H.S. = 1x5+2x6+3x7+....+k(k+4)+(k+1)(k+5) = k(k+1)(2k+13)/6+(k+1)(k+5) = k(k+1)(2k+13)/6+6(k+1)(k+5)/6 = (k+1)[k(2k+13)+6(k+5)]/6 = (k+1)[2k2+13k+6k+30]/6 = (k+1)[2k2+19k+30]/6 = (k+1)(k+2)(2k+15)/6 =R.H.S. So P(k+1) is true. By Mathematical Induction, 1x5+2x6+3x7+....+n(n +4)= n(n+1)(2n+13)/6 is true for all positive integer of n find 1x4+2x5+3x6+.....+n( n+3) 1x(5-1)+2x(6-1)+3x(7-1)+.....+n( n+4-1) =1x5 + 2x6 + 3x7 + ……+n(n+4) -1-2-3-….-n = n(n+1)(2n+13)/6 - n(n +1)/2 = n(n+1)(2n+13)/6 - 3n(n +1)/6 = n(n+1)[(2n+13) – 3]/6 = n(n+1)(2n+10)/6 = n(n+1)(n+5)/3 and 12+22+......+n2 = 1x1+1x4+2x2+2x4+3x3+3x4+…..+nxn+nx4–1x4–2x4–3x4-….-nx4 = 1x5+2x6+3x7+…..+n(n+4)–4(1+2+3+….+n) = n(n+1)(2n+13)/6–4n(n +1)/2 = n(n+1)(2n+13)/6–12n(n +1)/6 = n(n+1)[(2n+13)–12]/6 = n(n+1)[2n+13–12]/6 = n(n+1)(2n+1)/6

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