Prove by PMI

1) a+(a+d)+(a+2d)+ ......[a+(n-1)d] = n/2 [2a+(n-1)d], n E N.

2) n(n+1)(n+5) is divisible by 6 for all n E N.

3) 9 raised to n - 8n - 1 is a multiple of 64 for all n E N.

Assume p(k) is true:

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a+a+d+(a+2d)………..[a+(k-1)d]+ [a+kd ] =k/2[2a+(k-1)d]+ [a+kd ]

  =k(2a+kd-d)+2(a+kd)/2

  =2ak+kd-kd+2a+2kd/2

  =2ak+2a+kd+kd/2

  =2a(k+1)+kd(k+1)/2

  =(k+1)(2a+kd)/2

  =(k+1)/2(2a+kd)

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 It is same as our assumption where k is replaced by k+1. Therefore it is true for all natural numbers

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Assume p(k) is true
a+a+d+(a+2d)………..[a+(k-1)d]+ [a+kd ] =k/2[2a+(k-1)d]+ [a+kd ]

=k(2a+kd-d)+2(a+kd)/2

=2ak+kd-kd+2a+2kd/2

=2ak+2a+kd+kd/2

=2a(k+1)+kd(k+1)/2

=(k+1)(2a+kd)/2

=(k+1)/2(2a+kd)
It is same as our assumption where k is replaced by k+1. Therefore it is true for all natural numbers

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Hello Priyanka Verma,
I feel ur beautiful
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Hello Priyanka Verma,
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ha ha
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Here's the solution. Moreover one can prove that if the middle term is not divisible by 3 then the sum consecutive cubes is not divisible by 27 or any higher power of 3. Another thing to note is that the resulting sum relatively prime to any two of the numbers and share a greatest common factor (3) with the third.

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Priyanka verma can you please tell how you got a+kd
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Thanks Priyanka verma
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answer it

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Sorry to say but I could not understand will you plz explain..?..
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Give me answer

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Here the ans is I hope it will be useful to u

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What are you looking for?