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P(n)=1+4+7+.......+n(n+1)(n+2)= {n(n+1)(n+2)(n+3)}÷4Given for P(1) and P(m).Prove for P(m+1) |
| Answer» Let P(n) = {tex}1 \\cdot 2 \\cdot 3 + 2 \\cdot 3 \\cdot 4 + .... + n(n + 1)(n + 2){/tex}{tex} = \\frac{{n(n + 1)(n + 2)(n + 3)}}{4}{/tex}For n = 1{tex}P(1) = 1 \\times 2 \\times 3 = \\frac{{1 \\times 2 \\times 3 \\times 4}}{4} \\Rightarrow 6 = 6{/tex}{tex}\\therefore {/tex}\xa0P (1) is trueLet P(n) be true for n = k.{tex}\\therefore P(k) = 1 \\cdot 2 \\cdot 3 + 2 \\cdot 3 \\cdot 4 + ...{/tex}{tex} + (k + 1)(k + 2) = \\frac{{k(k + 1)(k + 2)(k + 3)}}{4}{/tex}\xa0... (i)For n = k + 1{tex}P(k + 1) = 1 \\cdot 2 \\cdot 3 + 2 \\cdot 3 \\cdot 4 + ... + {/tex}{tex}k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3){/tex}{tex} = \\frac{{k(k + 1)(k + 2)(k + 3)}}{4}{/tex}\xa0+ (k + 1) (k + 2) (k + 3) [Using (i)]{tex} = (k + 1)(k + 2)(k + 3)\\left[ {\\frac{{k + 4}}{4}} \\right]{/tex}{tex} = \\frac{{(k + 1)(k + 2)(k + 3)(k + 4)}}{4}{/tex}{tex}\\therefore {/tex}\xa0P(k + 1) is true.Thus P(k) is true {tex} \\Rightarrow {/tex}\xa0P (k + 1) is true.Hence by principle of mathematical induction, P(n) is true for all {tex}n \\in N{/tex}. | |