# Algebra precalculus

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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) + 1ig> \&=& 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\$.

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(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.

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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.

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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*.\$\$