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  • Prove the following by using the principle of mathematical induction for all n ∈ N: (2) 1^3+2^3+3^3+...+n^3=(n(n+1)/2)^2

Prove the following by using the principle of mathematical induction for all n\in N:

Q : 2        1^3+2^3+3^3+...+n^3=\left (\frac{n(n+1)}{2} \right )^2

Answers (1)

best_answer

Let the given statement be p(n) i.e.
p(n):1^3+2^3+3^3+...+n^3=\left (\frac{n(n+1)}{2} \right )^2
For n = 1  we have
p(1):1=\left (\frac{1(1+1)}{2} \right )^2= \left ( \frac{1(2)}{2} \right )^2=(1)^2=1    ,   which is true

For  n = k  we have
p(k):1^3+2^3+3^3+...+k^3=\left (\frac{k(k+1)}{2} \right )^2 \ \ \ \ \ \ \ \ \ \ - (i)   ,        Let's assume that this statement is true

Now,
For  n = k + 1  we have
p(k+1):1^3+2^3+3^3+...+(k+1)^3=1^3+2^3+3^3+...+k^3+(k+1)^3
                                                                                  =(1^3+2^3+3^3+...+k^3)+(k+1)^3
                                                                                  =\left ( \frac{k(k+1)}{2} \right )^2+(k+1)^3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
                                                                                  =\frac{k^2(k+1)^2+4(k+1)^3}{4}
                                                                                  =\frac{(k+1)^2(k^2+4(k+1))}{4}
                                                                                  =\frac{(k+1)^2(k^2+4k+4)}{4}
                                                                                  =\frac{(k+1)^2(k+2)^2}{4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because a^2+2ab+b^2=(a+b)^2)
                                                                                  =\left ( \frac{(k+1)(k+2)}{2} \right )^2
Thus,  p(k+1)  is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n)  is true for all natural numbers n

Posted by

Gautam harsolia

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