Mathematical Induction

1)1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6When n = 1 , LHS = 1^2 = 1(1+1)(2+1)/6 = RHSAssume that when n = k the statement is true , i.e.
1^2 + 2^2 + 3^2 + ... + k^2 = k(k+1)(2k+1)/6When n = k+1 ,
1^2 + 2^2 + 3^2 + ... + k^2 + (k+1)^2 = k(k+1)(2k+1)/6 + (k+1)^2= (k+1) [ k(2k+1)/6 + 6(k+1)/6 ]= (k+1) (2k^2 + 7k + 6)/6= (k+1) (2k+3)(k+2)/6= (k+1) (k+2) (2(k+1) +1) / 6When n = k is true , then n = k+1 is also true.
By Mathematical Induction , it is true for all positive integer.Putting n = 50 ,1^2 + 2^2 + 3^2 +.......... 50^2= (50)(51)(50*2 + 1)/6= 42925
2)1(2) + 2(2^2) + 3(2^3) + ... + n(2^n) = (n-1)[2^(n+1)] + 2When n = 1 , LHS = 1(2) = (1-1)[2^(1+1)] + 2 = RHS ,Assume that when n = k the statement is true , i.e.1(2) + 2(2^2) + 3(2^3) + ... + k(2^k) = (k-1)[2^(k+1)] + 2When n = k+1 ,1(2) + 2(2^2) + 3(2^3) + ... + k(2^k) + (k+1)[2^(k+1)] = (k-1)[2^(k+1)] + 2 + (k+1)[2^(k+1)] = (k-1 + k+1) [2^(k+1)] + 2= 2k [2^(k+1)] + 2= k [2^(k+2)] + 2When n = k is true , then n = k+1 is also true.
By Mathematical Induction , it is true for all positive integer.Putting n = 98 ,
1(2) + 2(2^2) + 3(2^3)+......98(2^98) = 97(2^99) + 2