many M.I

1. 1^2+3^2+...+(2n-1)^2 = (n(2n-1)(2n+1))/3 for all positive integer n

(1)(2(1)-1)(2(1)+1)/3 = 1 = 1^2

assume 1^2+3^2+...+(2k-1)^2 = k(2k-1)(2k+1)/3

consider

1^2+3^2+...+ (2k-1)^2 + (2k+1)^2

= k(2k-1)(2k+1)/3 + (2k+1)^2

= [ k(2k-1) + 3(2k+1) ] (2k+1)/3

= [ 2k^2 - k + 6k + 3 ] (2k+1)/3

= [ 2k^2 + 5k + 3 ] (2k+1)/3

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

= (k+1)(2(k+1)-1)(2(k+1)+1)/3

so, when case n = k is true, case n = k+1 is also true

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2. Tn=n^+n

proof T1+T2+...+Tn=1/3(n)( n+1)(n+2)

try yourself

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3.use the formula 1^2+2^2+...+n^2=1/6(n)(n+1)(2n+1)

find the sum 1x2+2x3+...+n(n+1)

try yourself

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4.proof 1x2+2x5+3x8+...+n(3n -1)=n^2(n+1)

try yourself

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5.proof 2n^3 + n is divisible by3

2(1)^3 + 1 = 3 is divisible by3

assume 2k^3 + k is divisible by3

2k^3 + k = 3m, m is integer

consider

2(k+1)^3 + k + 1

= 2(k^3 + 3k^2 + 3k + 1) + k + 1

= 2k^3 + 6k^2 + 7k + 3

= 2k^3 + k + 6k^2 + 6k + 3

= 3m + 6k^2 + 6k + 3

= 3(m + 2k^2 + 2k + 1)

because m + 2k^2 + 2k + 1 is integer, 2(k+1)^3 + k + 1 is divisible by3

so, when case n = k is true, case n = k+1 is also true

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6. 1^2-2^2+3^2-4^2+.....+(-1)^(n-1)n(n+1)/2

try yourself

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7.use 2(2)+3(2^2)+4(2^3)+...+(n+1)(2^n) = n(2^(n+1))

to show 1(2)+2(2^2)+3(2^3)+...+98(2^98) =97(2^99)+2

1(2)+2(2^2)+3(2^3)+...+98(2^98)

= 2(2)+3(2^2)+4(2^3)+...+(99)(2^98) - [ (2)+(2^2)+(2^3)+...+(2^98) ]

= (98)(2^(98+1)) - [ (2)+(2^2)+(2^3)+...+(2^98) ]

= (98)(2^(99)) - [ (2)(2^98 - 1)/(2-1) ]

= (98)(2^(99)) - [ (2)(2^98 - 1) ]

= (98)(2^(99)) - 2^99 + 2

= (97)(2^(99)) + 2

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8.proof n^3 - n + 3 is divisible by 3 for all...... n.

1^3 - 1 + 3 = 3 is divisible by 3

assume k^3 - k + 3 is divisible by 3

k^3 - k + 3 = 3m, m is integer

consider

(k+1)^3 - (k+1) + 3

= k^3 + 3k^2 + 3k + 1 - k - 1 + 3

= k^3 + 3k^2 + 2k + 3

= k^3 - k + 3 + 3k^2 + 3k

= 3m + 3k^2 + 3k

= 3(m + k^2 + k)

because m + k^2 + k is integer, (k+1)^3 - (k+1) + 3 is divisible by 3

so, when case n = k is true, case n = k+1 is also true

你都知多啦, 幫你做一 d, 其他自己試下做, 唔好咁懶