Gamitin ang Binomial Theorem upang mapalawak (x + 7) ^ 4 at ipahayag ang resulta sa pinasimple form?

Sagot:

# 2401 + 1372x + 294x ^ 2 + 28x ^ 3 + x ^ 4 #

Paliwanag:

Paggamit ng binomial teorama maaari naming ipahayag # (a + bx) ^ c # bilang isang pinalawak na hanay ng # x # mga tuntunin:

# (a + bx) ^ c = sum_ (n = 0) ^ c (c!) / (n! (c-n)!) a ^ (c-n) (bx) ^ n #

Narito, mayroon kami # (7 + x) ^ 4 #

Kaya, upang mapalawak ang ginagawa namin:

# (4!) / (0! (4-0)!) 7 ^ (4-0) x ^ 0 + (4!) / (1! (4-1)!) 7 ^ (4-1) x ^ 1 + (4!) / (2! (4-2)!) 7 ^ (4-2) x ^ 2 + (4!) / (3! (4-3)!) 7 ^ (4-3) x ^ 3 + (4!) / (4! (4-4)!) 7 ^ (4-4) x ^ 4 #

# (4!) / (0! (4-0)!) 7 ^ 4x ^ 0 + (4!) / (1! (4-1)!) 7 ^ 3x ^ 1 + (4!) / (2 (4-2)!) 7 ^ 2x ^ 2 + (4!) / (3! (4-3)!) 7x ^ 3 + (4!) / (4! (4-4)!) 7 ^ 0x ^ 4 #

# (4!) / (0! 4!) 7 ^ 4 + (4!) / (1! 3!) 7 ^ 3x + (4!) / (2! 2!) 7 ^ 2x ^ 2 + (4!) / (3! 1!) 7x ^ 3 + (4!) / (4! 0!) X ^ 4 #

# 7 ^ 4 + 4 (7) ^ 3x + 24/4 7 ^ 2x ^ 2 + 4 (7) x ^ 3 + x ^ 4 #

# 2401 + 1372x + 294x ^ 2 + 28x ^ 3 + x ^ 4 #

Gamitin ang DeMoivre's Theorem upang mahanap ang ikalabindalawa (ika-12) kapangyarihan ng kumplikadong numero, at isulat ang resulta sa karaniwang form?

(2 [cos ( frac { pi} {2}) + ako kasalanan ( frac { pi} {2})]) ^ {12} = 4096 Sa tingin ko ang tanong ay humihingi ng (2 [cos ( frac { pi} {2}) + i sin ( frac { pi} {2})]) ^ {12} gamit ang DeMoivre. (2 [cos ( frac { pi} {2}) + ako kasalanan ( frac { pi} {2})]) ^ {12} = 2 ^ {12} (cos (pi / 2) + i sin (pi / 2)) ^ 12 = 2 ^ {12} (cos (6 pi) + i sin (6pi)) = 2 ^ 12 (1 + 0 i) = 4096 Suriin: Hindi talaga kailangan ang DeMoivre ang isang ito: cos (pi / 2) + i sin (pi / 2) = 0 + 1i = ii ^ 12 = (i ^ 4) ^ 3 = 1 ^ 3 = 1 kaya kami ay naiwan sa 2 ^ }.

Paano mo ginagamit ang Binomial Theorem upang mapalawak (x + 1) ^ 4?

X ^ 4 + 4x ^ 3 + 6x ^ 2 + 4x + 1 Ang binomial theorem ay nagsasaad: (a + b) ^ 4 = a ^ 4 + 4a ^ 3b + 6a ^ 2b ^ 2 + 4ab ^ 3 + b ^ 4 dito, a = x at b = 1 Nakukuha namin ang: (x + 1) ^ 4 = x ^ 4 + 4x ^ 3 (1) + 6x ^ 2 (1) ^ 2 + 4x (1) ^ 3 + ^ 4 (x + 1) ^ 4 = x ^ 4 + 4x ^ 3 + 6x ^ 2 + 4x + 1

Paano mo ginagamit ang Binomial Theorem upang mapalawak (x-5) ^ 5?

(-5 + x) ^ 5 = -3125 + 3125x -1250x ^ 2 + 250x ^ 3-25x ^ 4 + x ^ 5 (a + bx) ^ n = sum_ (r = 0) ^ n ((n) (r)) a ^ (nr) (bx) ^ r = sum_ (r = 0) ^ n (n!) / x) ^ 5 = sum_ (r = 0) ^ 5 (5!) / (r! (5-r)!) (- 5) ^ (5-r) (5!) / (0! (5-0)!) (- 5) ^ (5-0) x ^ 0 + (5!) / (1! (5-1) 5-1) x ^ 1 + (5!) / (2! (5-2)!) (- 5) ^ (5-2) x ^ 2 + (5!) / (3! !) (- 5) ^ (5-3) x ^ 3 + (5!) / (4! (5-4)!) (- 5) ^ (5-4) x ^ 4 + (5!) / (5-5 (5-5)!) (- 5) ^ (5-5) x ^ 5 (-5 + x) ^ 5 = (5!) / (0! (5!) / (1! 4!) (- 5) ^ 4x + (5!) / (2! 3!) (- 5) ^ 3x ^ 2 + (5!) / (3! 5) ^ 2x ^ 3 + (5!) / (4! 1!) (- 5) x ^ 4 + (5!) / (5! 0!) X ^ 5 (-5 + x) ^ 5 = -5) ^ 5 + 5 (