# 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.

Xem thêm: Một Người Có 10 Đôi Giày Khác Nhau Và Trong Lúc Đi Du Lịch Vội Vã Lấy Ngẫu Nhiên 4 Chiếc  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.) 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)2igg)^2\$. Now consider \$displaystyle 1^3 + 2^3 + 3^3 + cdots + n^3 + (n + 1)^3 = igg(fracn(n+1)2igg)^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\$. 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 Site kiến thiết / logo sản phẩm © 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. Rev2022.12.2.43073