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