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Question

To find:
12+22+32...................+n2


  1. =n32

  2. >n32

  3. =n33

  4. >n33


Solution

The correct option is D

>n33


Let n=1
n32=12;    n33=13
Let n=2,12+22=5
n32=4;    n33=83
Let n=3,12+22+32=14
n32=272;    n33=9
Let n=4,12+22+32+42=30
n32=32;    n33=643
12+22+32++n2>n33
P(n):12+22+32++n2>n33
P(1) is true
Let P(k) be true.
12+22+32++k2>k33
12+22+32++k2+(k+1)2>k33+k2+2k+1
=k3+3k2+6k+33
=k3+3k2+3k+1+3k+23
=(k+1)33+k+23
12+22+32++k2+(k+1)2>(k+1)33
P(k+1) is true.
P(n) is true  nN

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