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Question

Solve : 13+23+33+...+n3=14n2(n+1)2

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Solution

P(n)=13+23+33+...+n3=(n(n+1)2)2
for n = 1
LHS = 13
RHS =(1(1+1)2)2=1
LHS = RHS
P(x) is true for n =1
P(k)
13+23+33+...+k3=(k(k+1)2)2
P(k+1)
13+23+...+k3+(k+1)3=(k(k+1)2)2+(k+1)3
=k2(k+1)222+(k+1)2
=k2(k+1)2+4(k+1)34
=(k+1)2(k+2)24
=((k+1)(k+2)2)2
This 13+2333+...+k3+(k+1)3=((k+1)(k+2)2)2 i.e;
P(k+1) is five whenever p(k) is the
By the principle of mathematical on distron
P(n) is true for n where n is natural number

1037468_1069294_ans_590f3421e2df49a18d9d34bb6c35a258.png

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