Răspuns:
Explicație pas cu pas:
b)
1/1·2=1/1-1/2
1/2·3=1/2-1/3
1/3·4=1/3-1/4
....................
1/(n-1)·n=1/(n-1)-1/n
1n(n+1)=1/n-1(n+1) daca adunam , in dreapta, termenii se reduc
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1/1·2+1/2·3+......+1/n(n+1)=1-1(n+1)=(n+1-1)/(n+1)=n/(n+1)
prin inductie
verificam n=1
1/1·2=1/(1+1) 1/2=1/2
admitem ca e adevarata relatia pentru n⇒ verificam ca va fi adevarata si pt n+1
1/2+1/2·3+.......+1/n(n+1)+1/(n+1)(n+2)=n/(n+1)+1/(n+1)(n+2)=1/(n+1)×[n+1/(n+2)]
1/(n+1)×[(n²+2n+1)/(n+2)=(n+1)²/(n+1)(n+2)=(n+1)/(n+2)
QED
c)1+3²+5²+....+(2n-1)²=n(4n²-1)/3
verificam pentru n=1
2-1=1(4-1)/3 ⇒1=3/3 adevarata
presupunem ca e adevarata pt n, verificam ca avem egalitate soi pt (n+1)
1²+3²+........+(2n-1)²+(2n+1)²=(n+1)[(4(n+1)²-1]/3
n(4n²-1)/3+(2n+1)²=[4n³-n+3(4n²+4n+1)]/3=(4n³-n+12n²+12n+3)/3=
=(4n³+12n²+11n+3)/3=(4n³+4n²+8n²+8n+3n+3)/3=(n+1)(4n²+8n+3)/3
4n²+8n+3=0
n1,2=-4±√16-12/4=-4±2/4 n1=-3/2 n2=-1/2
=(n+1)(2n+3)(2n+1)/3
partea dreapta
(n+1)[(4(n+1)²-1]/3=(n+1)[(2n+2-1)(2n+2+1)]/3=(n+1)(2n+1)(2n+3)/3
QED
