The Sum Of The Cubes Is The Square Of The Sum

     

I.H.: Assume that, for some $k in Bbb N$, $1^3 + 2^3 + ... + k^3 = (1 + 2 +...+ k)^2$.

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Want lớn show that $1^3 + 2^3 + ... + (k+1)^3 = (1 + 2 +...+ (k+1))^2$

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

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

$ = (1+2+...+k)^2 + (k+1)^3$ by I.H.

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Annnnd I"m stuck. Not sure how lớn proceed from here on.

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HINT: You want that last expression to turn out lớn be $ig(1+2+ldots+k+(k+1)ig)^2$, so you want $(k+1)^3$ to be equal to lớn the difference

$$ig(1+2+ldots+k+(k+1)ig)^2-(1+2+ldots+k)^2;.$$

That’s a difference of two squares, so you can factor it as

$$(k+1)Big(2(1+2+ldots+k)+(k+1)Big);. ag1$$

To show that $(1)$ is just a fancy way of writing $(k+1)^3$, you need to show that

$$2(1+2+ldots+k)+(k+1)=(k+1)^2;.$$

Do you know a simpler expression for $1+2+ldots+k$?

(Once you get the computational details worked out, you can arrange them more neatly than this; I wrote this specifically to lớn suggest a way khổng lồ proceed from where you got stuck.)


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Consider the case where $n = 1$. We have $1^3 = 1^2$. Now suppose $1^3 + 2^3 + 3^3 + cdots + n^3 = (1 + 2 + 3 + cdots + n)^2$ for some $n in ccevents.vnbb N$. Recall first that $displaystyle (1 + 2 + 3 + cdots + n) = fracn(n+1)2$ so we know $displaystyle 1^3 + 2^3 + 3^3 + cdots + n^3 = igg(fracn(n+1)2igg)^2$. Now consider $displaystyle 1^3 + 2^3 + 3^3 + cdots + n^3 + (n + 1)^3 = igg(fracn(n+1)2igg)^2 + (n+1)^3 = fracn^2 (n+1)^2 + 4(n+1)^34 = igg( frac(n+1)(n+2)2 igg)^2$. Hence, the statement holds for the $n + 1$ case. Thus by the principle of ccevents.vnematical induction $1^3 + 2^3 + 3^3 + cdots + n^3 = (1 + 2 + 3 + cdots + n)^2$ for each $n in ccevents.vnbb N$.


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IMHO, this fact is a coincidence; a better approach is khổng lồ prove the closed-form formula for both. As we know

$$ 1 + 2 + cdots + k = frack(k+1)2 $$

the corresponding claim khổng lồ prove is

$$ 1^3 + 2^3 + cdots + k^3 = frack^2(k+1)^24 $$


Prove $ fracsqrt22 + fracsqrt34 + fracsqrt46 + cdots + fracsqrtn+12n > fracsqrtn2 $ by induction
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