cho A=1.2.3 + 2.3.4 + … + n(n+1)(n+2). cmr 4A + 1 la 1 so chinh phuong

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1. Ta có: A=1.2.3 + 2.3.4 + … + n(n+1)(n+2)

   ⇔4A=1.2.3.4 + 2.3.4.4 + … + n(n+1)(n+2).4

   ⇔4A=1.2.3.4 + 2.3.4.(5-1) + … + n(n+1)(n+2)[(n+3)-(n-1)]

   ⇔4A=1.2.3.4 + 2.3.4.5-1.2.3.4 + … + n(n+1)(n+2)(n+3) – (n-1)n(n+1)(n+2)

   ⇔4A= n(n+1)(n+2)(n+3)

   ⇔4A=(n²+3n)(n²+3n+2)

   ⇔4A= (n²+3n)² + 2(n²+3n)

   ⇔4A+1=(n²+3n)² + 2(n²+3n) +1

   ⇔4A+1=(n²+3n+1)² là số chính phương. ⇒ đpcm

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2. Cách giải:

   $A=1.2.3+2.3.4+…….+n(n+1)(n+2)$

   $\to 4A=1.2.3.4+2.3.4.4+…..+n(n+1)(n+2).4$

   $\to 4A=1.2.3.4+2.3.4.(5-1)+…..+n(n+1)(n+2).[n+3-(n-1)]$

   $\to 4A=1.2.3.4+2.3.4.5-1.2.3.4+…..+n(n+1)(n+2)(n+3)-(n-1).n(n+1)(n+2)$

   $\to 4A=n(n+1)(n+2)(n+3)$

   $\to 4A+1=[(n+1)(n+2)][n(n+3)]+1$

   $\to 4A+1=(n^2+3n+2)(n^2+3n)+1$

   $\to 4A+1=(n^2+3n+1+1)(n^2+3n+1-1)+1$

   $\to 4A+1=(n^2+3n+1)^2-1+1$

   $\to 4A+1=(n^2+3n+1)^2$ là 1 số chính phương.

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