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

$$1^3+2^3+3^3+.......+n^3=(\dfrac{n(n+1)}{2})^2$$

Asked by Abhisek | 1 year ago |  67

#### 1 Answer

##### Solution :-

Let us denote the given equality by P(n), i.e.,

$$1^3+2^3+3^3+.......+n^3=(\dfrac{n(n+1)}{2})^2$$

For n=1,
L.H.S. 13=1

R.H.S.= $$( \dfrac{1(1+1)}{2})^2$$

= (1)2= 1

Therefore,P(n) is true for n=1.

Let us assume that P (k) is true for some positive integer k , i.e.,

$$1^3+2^3+3^3+.......+k^3=(\dfrac{k(k+1)}{2})^2$$ ..........(i)

Now, we have to prove that P(k+1) is also true.

Consider

$$1^3+2^3+3^3+.......+k^3(k+1)^3$$

$$( 1^3+2^3+3^3+.....+k^3)(k+1)^3$$

$$(\dfrac{k(k+1)}{2})^2+(k+1)^3$$......... [Using (i)]

$$(\dfrac{k^2(k+1)^2}{4})^2+(k+1)^3$$

$$\dfrac{k^2(k+1)^2+4(k+1)^3}{4}$$

$$\dfrac{(k+1)^2(k^2+4(k+1)}{4}$$

$$\dfrac{(k+1)^2(k+2)^2}{4}$$

$$\dfrac{(k+1)^2(k+1+1)^2}{4}$$

$$\dfrac{(k+1)(k+1+1)^2}{2}$$

Therefore,P (k+1)  holds whenever  P(k) holds.

Hence, the given equality is true for all natural numbers i.e., N by the principle of mathematical induction.

Answered by Pragya Singh | 1 year ago

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