URGENT!! VA ROG AJUTOR, DOAR SUBPUCTELE C SI D, VA ROOOG!!
Răspunsuri la întrebare
d) 1 * 4 + 4 * 7 + .... + (3n-2)(3n+1) = n(3n² + 3n - 2)
1. Etapa de verificare
p(1) : 1 ( 3 + 3 -2) = 1 *4 = 4
1 * 4 = 4
-> adevarat
2. Demonstrația
p(k)(1) -> p(k+1)(1)
p(k): 1 * 4 + 4* 7 + ... + (3k-2)(3k+1) = k(3k²+3k-2)
p(k+1) : 1 * 4 + .... (3k-2)(3k+1) + (3k+1)(3k+4) =( k+1)[3(k+1)² + 3k + 3 - 2]
3k³ + 3k² - 2k + 9k² + 12k + 3k + 4 = ( k + 1) [ 3(k² + 2k + 1) + 3k + 1)
3k³ + 12k² + 13k + 4 = (k + 1)( 3k² + 9k+ 4)
3k³ + 12k² + 13k + 4 = 3k³ + 9k² + 4k + 3k² + 9k + 4
3k³ + 12k² + 13k + 4 = 3k³ + 12k² + 13k + 4 Adevarat.
c)
1. Et de verificare
p(1) : 1 ( 16 + 36 + 11)/3 = 21
3 * 7 = 21
-> adevarat
2. Demonstrația
p(k)(1) -> p(k+1)(1)
p(k) : 3 * 7 + 7 * 11 + ... + (4k-1)(4k+3) = k(16k² + 36k + 11)/3 = (16k³ + 36k² + 11k)/3
p(k+1) : 3 * 7 + 7 * 11 + .... (4k + 3)(4k + 7) = (k+1)[16(k+1)² + 36k + 36 + 11) /3
= (k+1)[ 16K² + 32k + 16 + 36k + 36 + 11] / 3
= (k+1)[ 16K² + 68k + 63]
= 16k³ + 68k² + 63k + 16k² + 68k + 63
= 16k³ + 84k² + 131k + 63)/3
(4k+3)(4k+7) = 16k² + 28k + 12k + 21
16k² + 28k + 12k + 21 + (16k³ + 36k² + 11k)/3= (16k³ + 84k² + 131k + 63)/3
16k³
48k² + 84k + 36k + 63 + 16k³ + 36k² + 11k = 16k³ + 84k² + 131k + 63
16k³ + 131k + 84k² + 63 = 16k³ + 131k + 84k² + 63 -> Adevarat.