ALGEBRA PRECALCULUS

  -  
Prove sầu that for all integers $n$, $n geq 1$, $$1 + 3 + 5 + cdots + (2n - 1) = ncdot n$$

How would I go about proving this?


*

*

(1) Proof using the "method of gauss":

$$egineqnarray* 2(1 + 3+ 5 + ldots (2n-1)) &=&ig< 1 + (2n-1)ig> + ig<3 + (2n-3)ig> + ldots + ig< (2n-1) + 1ig> \&=& underbrace2n + 2n + ldots + 2n_ ext$n$ times \&= &2n(n). endeqnarray*$$ Therefore it follows that $(1 + 3 + ldots (2n-1)) = n imes n.$

(2) Proof by induction: Let $P(n)$ be the statement:

"For all positive integers $n$, $1 + 3 + ldots + (2n-1) = n^2$."

For $n=1$ clearly $(2 cdot 1) - 1 = 1cdot 1$ so that $P(1)$ is true. So suppose the result holds for $n=k$, i.e. $P(k)$ is true. Since $P(k)$ is true, this means that we have the following equality when $n=k$:

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

If you add $2k+1$ to lớn both sides of this equation, you get that

$$ egineqnarray* 1+ 3 + ldots +(2k-1) + (2k+1) &=và kcdot k + (2k+1) \&=& k^2 + 2k + 1 \&=& (k+1)(k+1) endeqnarray*$$

showing that the result holds for $n=k+1$, i.e. $P(k+1)$ is true. Therefore by the Principle of pgdtxhoangmai.edu.vnematical Induction, $1 + 3 + ldots + (2n-1) = ncdot n$ for all positive integers $n$.

Bạn đang xem: Algebra precalculus

(3) Suppose you only knew that the sum of $n$ consecutive sầu integers is $n(n+1)/2$.Then

$$egineqnarray*1 +2 + ldots + 2n &=và frac2n(2n+1)2\implies 1 + 3 + ldots + (2n-1) &=& n(2n+1) - 2(1 + ldots + n) \&=và 2n^2 + n - 2left(fracn(n+1)2 ight) \&=& 2n^2 + n - n^2 - n \&=& n^2.

Xem thêm: Tại Sao Không Làm Pittông Vừa Khít Với Xilanh Để Không Phải Sử Dụng Xecmăng

endeqnarray*$$

(4) Proof using telescoping sums (idea by Bill Dubuque):

$$egineqnarray* sum_k=1^n 2k-1 &=& sum_k=1^n k^2 - (k-1)^2 \&=& (1 - 0) + (2^2 -1 ) + ldots + n^2 - (n-1)^ 2\&=và 1 + (-1 + 2^2) + (-2^2 + 3^2) + ldots + (-(n-2)^2 + (n-1)^2) + (-(n-1)^2 + n^2)) \&=& n^2.endeqnarray*$$

(5) Proof by method of differences (brute force): Let $a_n = sum_k=1^n 2k-1$. Then we see that $a_1 =1 $, $a_2 = 4$, $a_3 = 9$, etc.

Xem thêm: Bài Văn Kể Về Gia Đình Em Lớp 6, Kể Về Gia Đình Mình Lớp 6

Now we look at the first differences $4 - 1 = 3$, $9- 4 = 5$, $16 - 9 = 7$, etc. Then when we look at the second difference, notice that it is constant: $5-3 = 2$, $7- 5= 2$, etc so we conjecture that

$$sum_k=1^n 2k - 1 = an^2 + bn + c$$

where $a,b,c$ are constants to be determined. Plugging in $n = 1,2,3$ gives us a $3 imes 3$ linear system khổng lồ solve sầu, namely the linear system

$$left<eginarrayccc 1 & 1 và 1 \ 4 và 2 & 1 \ 9 và 3 & 1 endarray ight>left<eginarrayc a \ b \ c endarray ight> = left<eginarrayc 1 \ 4 \ 9 endarray ight>.$$

The determinant of the coefficient matrix is

$$egineqnarray* 1(2-3) - 1(4-9) + 1(12 - 81) &=& -1 + 5 - 69 \& eq& 0 endeqnarray*$$ so we have a quality solution. It is easy lớn see that $a = 1, b= 0, c=0$ is a solution lớn the linear system above. By the previous line, it is the only solution so we are done.

(6) Proof by a direct bash: Suppose you only knew that the sum of the first $n$ integers is $n(n+1)/2$. Then

$$egineqnarray* sum_k=1^n 2k - 1 &=và igg(2sum_k=1^n k igg) - sum_k=1^n 1 \&=& 2left(fracn(n+1)2 ight) - n\&=và n^2 + n -n\&=và n^2endeqnarray*.$$