Cap 2 2 - Resolução de Exercicios

Prévia do material em texto

1+5+9+...+(4n -3) = n(2n-1)
1. p(n): n(2n-1) 
 p(1): 1(2.1-1) = 1 
2. p(k): 1+5+9+...+(4k -3)=k(2k-1)
3. p(k+1): (4(k+1)-3) =k+1(2(k+1)-1)
(4(k+1) -3)+(k(2k-1)) =k+1(2(k+1)-1)
(4(k+1) -3)+(k(2k-1)) = 2k²+3k+1
2k²+3k+1 = 2k²+3k+1
4+10+16+...+(6n-2) =n(3n+1)
1. p(n): n(3n+1) 
 p(1): 1(3.1+1) = 4 
2. p(k):4+10+16+...+(6k-2)=k(3k+1)
3. p(k+1): (6(k+1))-2)=k+1(3(k+1)+1)
((6k+1)-2)+k(3k+1) = k+1(3(k+1)+1)
(6k+4)+ 3k²+k = (k+1)(3k+4)
3k²+7k+4 = 3k²+7k+4 
1²+3²+...+(2n-1)² = (n(2n-1)(2n+1)) / 3
1. p(n): (2n-1)² = (n(2n-1)(2n+1)) / 3 
 p(1): 1² = 3/3 → 1 = 1 
2. p(k): 1²+3²+...+(2k-1)² = (k(2k-1)(2k+1)) / 3
3. p(k+1): (2(k+1)-1)² = ((k+1)(2(k+1)-1)(2(k+1)+1)) / 3
(2(k+1)-1)² + (k(2k-1)(2k+1)) / 3 = ((k+1)(2(k+1)-1)(2(k+1)+1)) / 3
4k³+ 12k²+11k+3 / 3 = (k+1)(2k+1)(2k+3)/3
4k³+ 12k²+11k+3 / 3 = 4k³+ 12k²+11k+3 / 3
 2+6+18+...+2*3^(n-1) = (3^n)-1 
1. p(n): (3^n)-1 → p(1): (3^1)-1 = 2
2. p(k): 2.3^(k-1) = (3^k)-1
3. p(k+1): 2.3^((k+1)-1) = (3^(k+1))-1
 2.3^(k) + (3^k)-1 = (3^(k+1))-1
6^(k) + (3^k)-1 = (3^(k+1))-1
(9^k)- 1 = (3^(k+1))-1 
3¹. 3^k -2 = (3^(k+1))-1 
(3^(k+1))-1 = (3^(k+1))-1 
1.1!+2.2!+3.3!+...+n.n! = (n+1)! - 1 
1. p(n): n.n! = (n+1)! - 1
 p(1): 1.1! = (1+1)! - 1
2. p(k): k.k! = (k+1)! - 1
3. p(k+1):(k+1).(k+1)! = ((k+1)+1)!- 1
(k+1)(k+1)!+(k+1)! - 1 = (k+2)!- 1
(k+1)!+(k+!)(1+k+1)-1 = (k+2)!- 1
(k+1)!+(k+2)-1 = (k+2)!- 1
(k+2)! -1 = (k+2)-1
1/(1.2)+ 1/(2.3)+1/(3.4)+...+1/n(n+1) = n/n+1
1. p(n): 1/n(n+1) = n/n+1
 p(1): 1/1(1+1) = 1/1+1 → ½ = ½
2. p(k): 1/(1.2)+ 1/(2.3)+1/(3.4)+...+1/k(k+1) = k/k+1
3. p(k+1): 1/(k+1)((k+1)+1) = (k+1)/((k+1)+1)
1/(k+1)((k+2)+ k/k+1 = (k+1)/((k+2)
[(k(k+2)+1)/(k+2)] . (k/(k+1)) 
(k²+2k+1)/(k+2)(k+1) = (k+1)/(k+2)
 n² > n+1 para n >= 2
1. p(n) = n² > n+1
 P(2) = 2² > 2+1 → 4 > 3
2. p(k): k² > k+1
3. p(k+1): (k+1)² > (k+1)+1
(k+1)² > k+2
(k+1)(k+1) > k+2 → k²+2k+1 > k+2
(k+1)+2k+1 > k+2
3k+2 >k+2
n = 4.a + 5.b ; n>=12 | 4 ,5 qtd selos
1. p(n): n = 4.a + 5.b
 12 = 4.3 + 5.0
2. p(k): k = 4.a + 5.b 
3. p(k+1) r = 4.a + 5.b; 12<= r <=k
(k-3) = (4.a+5.b)+4
(k+1)= 4a+5b+4
k+1 = 4(a+1)+5b
(1²-2²)+(3²-4²)+...+((-1)^(n+1)) n² = (((-1)^(n+1))(n)(n+1)) / 2
1. p(n): ((-1)^(n+1)) n² = (((-1)^(n+1))(n)(n+1)) / 2
 p(1): 1.1 = 2/2 → 1 =1 
2. p(k):((-1)^(k+1)) k² = (((-1)^(k+1))(k)(k+1)) / 2
3. p(k+1): ((-1)^((k+1)+1)) (k+1)² = (((-1)^((k+1)+1))(k+1)
((k+1)+1)) / 2
((-1)^((k+2) (k+1)²+ ((-1)^(k+1))(k)(k+1) / 2= (-1)^((k+2)(k+1)
(k+2) / 2
(-1)^((k+2)(k+1)(k+2) / 2 = (-1)^((k+2)(k+1)(k+2) / 2
2^(3n) – 1 é divisivel por 7 
1. p(n): 2^(3n) – 1
 p(1): 2^(3) – 1 = 8 -1 = 7 
2. p(k): 2^(3k) – 1 = 7.a 
3. p(k+1): 2^(3(k+1)) – 1 = 7.b 
2^(3k+3) – 1 = 2^(3k).2³+1
a+ar+ar²+...+ar^(n-1)=(a - ar^n)/ 1-r PG
1. p(1) = a 
 (a-ar)/1-r = a(1-r)/1-r = a
2. p(k): ar^(k-1) = (a – ar^k) / 1-r
3. p(k+1): ar^((k+1)-1) = (a-ar^(k+1)) / 1-r
 ar^((k)+ (a - ar^k)/ 1-r = (a-ar^(k+1)) / 1-r
((1-r).ar^((k)+(a - ar^k))/ 1-r = (a-ar^(k+1)) / 1-r
(a - ar^(k+1)) / 1-r = (a - ar^(k+1)) / 1-r