proof that p(1)+p(2)+...+p(n)=n^3 (please help)

Question

P(x) function polynomial 2nd degree P(1)=1 P(2)=7 P(3)=19

we have natural integer n ⁄ n≥1

proof that P(1)+P(2)+...+P(n)=n³

Help please

Answer 1

I. We know that P(1) =1, P(2) = 7 and P(3) =19. For a polynomial of second degree:

P(n) = a n² + b n +c:

P(1) = a 1² +b 1 + c = 1

P(2) = a 2² + b 2 + c =7

P(3) = a 3² + b 3 + c = 19.

The system looks like

a + b + c = 1

4a +2b + c = 7

9a +3b + c =19.

You can use any method to solve this system of equations. We find that

a = 3, b = -3, c = 1.

The polynomial:

P(n) = 3n² - 3n +1.

II. Prove that

Σ_n=1ⁿ P(n) = n³.

This part is very nicely done using induction.

Induction part 1. We check that the statement is true for the first natural number n=1.

Σ_n=1¹ P(n) = 1³.

We calculate:

Σ_n=1¹ P(n) = P(1) = 3 ⋅ 1² -3 ⋅ 1 + 1 = 1

1 = 1³ - TRUE.

We proved that if n = 1, the sum indeed equals to 1³.

Part 2. We will make an assumption. We say that the statement is true for some natural number, say k. That number k may be 2, or 1000 or 2561 or any other natural number.

Under our assumption we have the sum of the first polynomials to be k³. As a mathematical statement, it would look like

Assumption

Σ_i=1^k P(i) = k³

We will prove that

Σ_i=1^k+1 P(i) = (k +1)³.

Let work of the sum when we have k + 1 members. We can divide the sum into two components, first k and the last member:

Σ_i=1^k+1 P(i) = [Σ_i=1^k P(i)] + P(k+1).

In the above expression, the sum in the brackets is equal to k³ (remember our assumption above) and the second addend is P(k+1) = 3(k+1)²- 3(k+1) + 1. Now, we substitute these into the statement above:

Σ_i=1^k+1 P(i) = [k³] + 3(k+1)² - 3(k+1) +1

and simplify:

= k³ +3(k² +2k + 1) - 3k - 3 +1 .... evaluate (k+1)² and open the second set of parenthesis

= k³ + 3k² + 6k + 3 - 3k - 3 + 1 .... open the remaining set of parenthesis

= k³ + 3k² +3k + 1 ......................... combine like terms

= (k+1)³ .......................................... recognize the sum above is (k+1)³.

Working on the above identity, we saw that

Σ_i=1^k+1 P(i) = (k+1)³. TRUE!.

This concludes the induction proof.

Additional notes on induction proof:

In general,

1) We check a statement is true for some natural number N₀.

2a) We check that the statement is true for N_o + 1;

2b) We check that the statement is true for N_o + 2;

2c) We check that the statement is true for N_o + 3;

...

All the checks above are summarized in the two steps we performed above - Part 1 and Part 2. In the summary, I used 1) for Part 1 and 2a, 2b, 2c, ... for Part 2.