A-Maths (mathematical induction )**急

1(2) + 2(3) + ... + n (n + 1)
= 1(1+1) + 2(2+1) + 3(3+1) + ... + n(n+1)
=( 1(1) + 2*2 + 3*3 + ... + n*n) + (1*1 + 2*1 + 3*1 + ... + n*1)
= 1² + 2² + 3² + ... + n² + (1+2+3+...+n)
= 1/6*n(n+1)(2n+1) + (n/2*(n+1)).....{(1+2+3+...+n) is an Arithmetic Series}
= 1/6*n(n+1)[2n+1 + 3]
= 1/6*n(n+1)(2n+4)
= 1/3*n(n+1)(n+2)

As you say you just don't know part b only, I only work out part b.

b) 1(2) + 2(3) + ... + n (n + 1)

= (1)(1+1) + (2)(2+1) + ... + n(n+1)

= (1)(1)+1+(2)(2)+2+...+(n)(n)+n

= (1)^2+(2)^2+...+(n)^2+1+2+...+n

By the proved formula 1^2 + 2^2 + ... + n^2 = 1/6 n (n + 1) (2n + 1) and 1+2+...+n=1/2n(n+1),

(1)^2+(2)^2+...+(n)^2+1+2+...+n

= 1/6 n (n + 1) (2n + 1)+1/2n(n+1)

= 1/6n(n+1){(2n+1)+3}

= 1/6n(n+1)(2n+4)

= 1/3n(n+1)(n+2)

b)1(2) + 2(3) + ... + n (n + 1)
佢地既特點係.........n^2+n [[[[n (n + 1) ,n X 入去]]]]
係(a)你prove 左1^2 + 2^2 + ... + n^2 係= 1/6 n (n + 1) (2n + 1)
b放去a
1(2) + 2(3) + ... + n (n + 1)
=[1^2+1] + [2^2+2] + ... + n^2+n
= (1/6 n (n + 1) (2n + 1) )+n

我記得我讀a.math 呀sir 教睇條formula睇+n^2 果位就得.....係b part搵返 a part 有既野..禁就用得返a part 條formula