The answer process is as follows:

an = n？

Sn = 1？ + 2? + 3? + .+ n？ = n(n+ 1)(2n+ 1)/6

Inductive proof:

N = 1,/kloc-0 /× (1+1)× (2×1+1)/6 = 6/6 =1,and the summation formula is correct.

Let n = k and Sk = 1? + 2? + 3? + .+ k？ = k(k+ 1)(2k+ 1)/6 holds.

s(k+ 1)= k(k+ 1)(2k+ 1)/6+(k+ 1)？

=(k+ 1)[k(2k+ 1)/6+(k+ 1)]

=(k+ 1)[k(2k+ 1)+6k+6]/6

= (k+ 1)[2k？ +7k+6]/6

=(k+ 1)[(k+2)(2k+3)/6

=(k+ 1)[(k+ 1)+ 1][2(k+ 1)+ 1]/6

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Extended data:

Correlation formula:

( 1)(a+b)？ =a？ +3a？ b+3ab？ +b？

(2) a? +b？ =a？ +a？ b-a？ b+b？ =a？ (a+b)-b(a？ -B? )=a？ (a+b)-b(a+b)(a-b)

=(a+b)[a？ -b(a-b)]=(a+b)(a？ -ab+b？ )

(3) a? -B? =a？ -a？ b+a？ b-b？ =a？ (a-b)+b(a？ -B? )=a？ (a-b)+b(a+b)(a-b)

=(a-b)[a？ +b(a+b)]=(a-b)(a？ +ab+b？ )

(4) (A-B)? =a？ -3a？ b+3ab？ -B?

(a-b)? =(a-b)(a-b)？ =(a-b)(a？ -2ab+b？ )=a？ -3a？ b+3ab？ -B?

The simplest and most common mathematical induction is to prove that a proposition holds when n is equal to any natural number. Proof is divided into the following two steps:

1. Prove that the proposition holds when n= 1.

2. Assuming that the proposition holds when n=m, it can be deduced that the proposition also holds when n=m+ 1. (m stands for any natural number)

The principle of this method lies in: first, prove that the proposition is valid at a certain initial value, and then prove that the process from one value to the next is valid. When these two points are proved, then any value can be deduced by repeatedly using this method. It may be easier to understand this method by thinking of it as domino effect.