By the principle of Mathematical induction 1³+2³+3³+.........+n³=n²(n+1)²/4

For n = 1, L.H.S = 1^3 =1. R.H.S = n^2(n+1)^2 /4 = 1^2(1+1)^2/4 = 1×(2)^2/4 = 1×4/4 =1. L.H.S = R.H.S. Therefore the result is true for n=1. For n= k Assume that the result is true for n=k. 1^3 + 2^3 + 3^3 + ...............n ^3 = n^2(n+1)^2/4 put n = k 1^3 + 2^3 + 3^3 + ........k^3 = k^2(k+1)^2/4 name it as eqn. 1 FOR n = k+1 LHS => 1^3 + 2^3 + 3^3 + ........ k^3 + (k+1)^3 => we have found the value upto k^3 in eqn. 1. So we will use it now................... k^2(k+1)^2/4 + (k+1)^3 => (k+1)^2 [k^2 /4 + k+1] ........ Taking LCM in bracket we will have........ (k+1)^2 [ k^2 + 4k + 4/ 4].................. On taking roots of the eqn in bracket we will have (k+1)^2 (k+2)^2/4 RHS => (k+1)^2 (k+1+1)^2/4 => (k+1)^2 (k+2)^2/4.............. LHS = RHS. Therefore the result is true for n = k+1. By PMI, the result is true for every natural number.
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1³+2³+3³+...+n³=¼n²(n+1)²
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