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Najlepsza Odpowiedź!
2009-12-03T21:20:11+01:00
A)
1. n=1
L=1²=1
P=1(1+1)(2*1+1)/6=6/6=1
L=P
2.
Założenie 1²+2²+3²+...+n²=n(n+1)(2n+1)/6
Teza 1²+2²+3²+...+(n+1)²=(n+1)(n+2)(2n+3)/6

L=1²+2²+3²+...+n²+(n+1)²=n(n+1)(2n+1)/6 +(n+1)²=
=n(n+1)(2n+1)+6(n+1)²/6=(n+1)[n(2n+1)+6(n+1)]/6=
=(n+1)[2n²+n+6n+6)]/6=(n+1)[2n²+4n+3n+6)]/6=
=(n+1)[2n(n+2)+3(n+2)]/6=(n+1)(n+2)(2n+3)/6=P

b)
1. n=1
L=1³=1
P=[1(1+1)/2]²=1²=1
L=P
2.∀n≥1
Założenie 1³+2³+3³+...+n³=[n(n+1)/2]²
Teza 1³+2³+3³+...+(n+1)³=[(n+1)(n+2)/2]²

L=1³+2³+3³+...+(n+1)³=1³+2³+3³+...+n³+(n+1)³=
=[n(n+1)/2]²+(n+1)³={[n(n+1)]²/4}+(n+1)³=
={[n(n+1)]²+4(n+1)³}/4={(n+1)²[n²+4(n+1)]}/4=
={(n+1)²[n²+4n+4)]}/4=[(n+1)²(n+2)²]/4=[(n+1)(n+2)/2]²=P
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