Question

folosind metoda inductiei matemafice demonstreaza
3+11+19+.....+(8n-5)=n(4n-1).n€N*

Answer 1

3+11+19+...+(8n-5)=n(4n-1)
Etapa verificarii : P(1): 8-5=1(4-1)= > 3=3 (Adevarat)
Presupunem ca (k) e adv.si dem .P(k+1)
P(k): 3+11+19+...+(8k-5)=k(4k-1)
P(k+1): 3+11+19+...(8k-5)+8(k+1)-5=(k+1)[4(k+1)-1]
k(4k-1)+8k+3=(k+1)(4k+3)
4k²-k+8k+3=4k²+3k+4k+3
4k²+7k+3=4k²+7k+3 (A)
0=0 (A)
Conform principiului ind.matematice, P(n) adevarata.

Answer 2

poza
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