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Question

Solve : 12+22+32+...+n2=16n(n+1)(2n+1)

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Solution

Let p(n)=12+22+32+....+n2=n(n+1)(2n+1)6
for n=1
LHS=12=1
RHS=(1)(1+1)(2×1+1)6=1×2×36=1
LHS = RHS
P(n) is true for n=1
Assume that P(k) is true
12+22+32+...+k2=k(k+1)(2k+1)6
we will prove P(k+1) is true
12+22+32+....+(k+1)2=k(k+1)(2k+1)6+(k+1)2
=k(k+1)(2k+1)+6(k+1)26
=(k+1)(k(2k+1)+6(k+1))6
=(k+1)(2k2+k+6k+1)6
=(k+1)(2k2+7k+1)6
=(k+1)(k+2)(2k+3)6
Thus 12+22+32+...+(k+1)2=(k+1)(k+2)(2k+3)6
P(k+1) is true when p(k) is true
BY principle of mathematical induction , P(n) true for n where n is a natural number

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