# Prove by PMI1) 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. • -8
Priyanka verma can you please tell how you got a+kd
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Thanks Priyanka verma
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Sorry to say but I could not understand will you plz explain..?..
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Here the ans is I hope it will be useful to u • 25 • 0
Hi!

P.M.I is the abbreviation for Principle of mathematical induction.
Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true of all natural number.

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Cheers!
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