Câu hỏi của Contrim Đẹptrai

Question

giúp mik vs các cậu ơi mik sắp phải nộp r

help me

Tính B=1.2.3+2.3.4+...+(n-1)n(n+1)

cảm ơn mn

Diệu Huyền · Accepted Answer

B=1.2.3+2.3.4+.........+(n−1)n(n+1)B=1.2.3+2.3.4+.........+(n−1)n(n+1)

⇔4B=1.2.3.4+2.3.4.4+........+(n−1)n(n+1).4⇔4B=1.2.3.4+2.3.4.4+........+(n−1)n(n+1).4

⇔4B=(4−0).1.2.3+(5−1).2.3.4+.........+[(n+2)−(n−2)](n−1)n(n+1)⇔4B=(4−0).1.2.3+(5−1).2.3.4+.........+[(n+2)−(n−2)](n−1)n(n+1)

⇔4B=1.2.3.4−0.1.2.3+2.3.4.5−1.2.3.4+.......+(n−1)n(n+1)(n+2)(n+3)−(n−2)(n−1)n(n+1)⇔4B=1.2.3.4−0.1.2.3+2.3.4.5−1.2.3.4+.......+(n−1)n(n+1)(n+2)(n+3)−(n−2)(n−1)n(n+1)

⇔4B=(n−1)n(n+1)(n+2)⇔4B=(n−1)n(n+1)(n+2)

⇔B=(n−1)n(n+1)(n+2)4Câu trả lời: B=1.2.3+2.3.4+.........+(n−1)n(n+1)B=1.2.3+2.3.4+.........+(n−1)n(n+1)

⇔4B=1.2.3.4+2.3.4.4+........+(n−1)n(n+1).4⇔4B=1.2.3.4+2.3.4.4+........+(n−1)n(n+1).4

⇔4B=(4−0).1.2.3+(5−1).2.3.4+.........+[(n+2)−(n−2)](n−1)n(n+1)⇔4B=(4−0).1.2.3+(5−1).2.3.4+.........+[(n+2)−(n−2)](n−1)n(n+1)

⇔4B=1.2.3.4−0.1.2.3+2.3.4.5−1.2.3.4+.......+(n−1)n(n+1)(n+2)(n+3)−(n−2)(n−1)n(n+1)⇔4B=1.2.3.4−0.1.2.3+2.3.4.5−1.2.3.4+.......+(n−1)n(n+1)(n+2)(n+3)−(n−2)(n−1)n(n+1)

⇔4B=(n−1)n(n+1)(n+2)⇔4B=(n−1)n(n+1)(n+2)

⇔B=(n−1)n(n+1)(n+2)4

bé Bắp · Answer

Băng Băng 2k6 giúp mik lm ik, mik bậnCâu trả lời: Băng Băng 2k6 giúp mik lm ik, mik bận

Violympic toán 7

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