f.4 m.i. 1 question

a)When n = 1 , L.H.S. = 1³ - 1 = 0 , R.H.S. = 1(1+2)(1²-1)/4 = 0Assuming the proposition is true for n = k , i.e. (1³ - 1) + (2³ - 2) + (3³ - 3) + ... + (k³ - k) = k(k + 2)(k² - 1) / 4When n = k+1 ,(1³ - 1) + (2³ - 2) + (3³ - 3) + ... + (k³ - k) + ( (k+1)³ - (k+1) ) = k(k + 2)(k² - 1) / 4 + ( (k+1)³ - (k+1) ) = [ k(k + 2)(k² - 1) + 4(k+1)³ - 4(k+1) ] / 4= (k+1) [k(k+2)(k-1) + 4(k+1)² - 4] / 4= (k+1) [k(k+2)(k-1) + 4k(k+2)] / 4= (k+1) k(k+2)[(k-1) + 4] / 4= (k+1)(k+3) k(k+2) / 4= (k+1)(k+1 + 2) (k+1 -1)(k+1 + 1) / 4= (k+1)(k+1 + 2)( (k+1)² - 1 ) / 4The proposition is true for n = k+1.Therefore , by the principle of mathematical induction , the proposition is true for all positive integers n.
b)(1)(2)(3) + (2)(3)(4) + (3)(4)(5) + ... + (n)(n+1)(n+2)= (2-1)(2)(2+1) + (3-1)(3)(3+1) + (4-1)(4)(4+1) + ... + (n+1 - 1)(n+1)(n+1 + 1)= 2(2²-1) + 3(3²-1) + 4(4²-1) + ... + (n+1)((n+1)² - 1)= (2³ - 2) + (3³ - 3) + (4³ - 4) + ... + ( (n+1)³ - (n+1) ) = (1³ - 1) + (2³ - 2) + (3³ - 3) + (4³ - 4) + ... + ( (n+1)³ - (n+1) ) ... since 1³ - 1 = 0= (n+1)(n+1 + 2)( (n+1)² - 1 ) / 4 ... by the result of part a)= n (n+1) (n+2) (n+3) / 4