Question

Let S(n)=1+2+···+ n be the sum of the first n natural numbers and let C(n)=1 3 +2 3 + ···+ n3 be the sum of the first n cubes. Prove the following equalitiesby inductionon n, to arriveat thecurious conclusionthatC(n)=S2(n) for every

Answer 1

Answer:

Step-by-step explanation:

We have

$S_n =\frac{n(n+1)}{2} \\C_n = (\frac{n(n+1)}{2})^2$

To prove that $C_n = S_n^2$ by induction

Let P(n) be that statement for n

n=1 gives left side =1 =right side

True for n=1

Assume that P(k) is true

i.e. $S_k^2 = C_k$

We have to prove P(k+1) is true assuming P(k)

LHS of $P(k+1) = (\frac{(k+1)(k+2)}{2} )^2$

RHS = $C_k+(k+1)^3\\= (\frac{k(k+1)}{2} )^2+(k+1)^3$

since by adding last term with Ck gives the sum

For simplification take (k+1) square outside

RHS = $\frac{(k+1)^2}{4} [k^2+4(k+1)]\\=\frac{(k+1)^2}{4}(k+2)^2\\=LHS$

Thus proved by mathematical induction