# Thread: induction of series P1

1. ## induction of series P1

Could someone help me with these induction questions? I only need the bit on proving
n=k+1 using the assumption n=k
Prove:
1^3 + 2^3 + 3^3 +....+n^3 = (1+2+3+...+n)^2 for n = 1,2,3...

1-2+3-4...+n(-1)^(n+1) = 1/4[1 - (2n+1)(-1)^n] for n = 1,2,3...

5/(1.2.3) + 6/(2.3.4) +7/(3.4.5) +...+(n+4)/n(n+1)(n+2) = [n(3n+7]/[2(n+1)(n+2)]

Thank you ! =)

2. Originally Posted by christina
Could someone help me with these induction questions? I only need the bit on proving
n=k+1 using the assumption n=k
Prove:
1^3 + 2^3 + 3^3 +....+n^3 = (1+2+3+...+n)^2 for n = 1,2,3...

1-2+3-4...+n(-1)^(n+1) = 1/4[1 - (2n+1)(-1)^n] for n = 1,2,3...

5/(1.2.3) + 6/(2.3.4) +7/(3.4.5) +...+(n+4)/n(n+1)(n+2) = [n(3n+7]/[2(n+1)(n+2)]

Thank you ! =)
Hello Christina,

For number 1......

If.... $\displaystyle 1^3+2^3+3^3+....+n^3=(1+2+3+....n)^2$

Then.... $\displaystyle 1^3+2^3+3^3+...+n^3+(n+1)^3=(1+2+3+...+n+[n+1])^2$

Proof:

$\displaystyle (1+2+3+...+n)^2+(n+1)^3=[1+2+3+...+n+(n+1)]^2$ ?

$\displaystyle (1+2+3+...+n)^2+(n+1)^3=[(1+2+3+...+n)+(n+1)][(1+2+3+...+n)+(n+1)]$ ?

$\displaystyle (1+2+..+n)^2+(n+1)^3=(1+2+..+n)^2+2(n+1)(1+2+..+n) +(n+1)^2$ ?

$\displaystyle (n+1)^3=(n+1)(n+1)^2=(n+1)\left(\sum_{i=1}^ni+(n+1 )\right)$ ?

$\displaystyle (n+1)^2=2\frac{n(n+1)}{2}+n+1=(n+1)(n+1)$ ?

True.

Now you only need prove it for n=1,
since the inductive step has proven that...
if the formula is true for n=1, then it's true for n=2, 3, 4, 5, 6....

3. Originally Posted by christina
Prove:

1-2+3-4...+n(-1)^(n+1) = 1/4[1 - (2n+1)(-1)^n] for n = 1,2,3...
$\displaystyle 1-2+3-4+...+n(-1)^{n+1}=\frac{1}{4}\left(1-(2n+1)(-1)^n\right),\ n=1,\ 2,\ 3...$ ?

$\displaystyle \left(1-2+3-4+...n(-1)^{n+1}+(n+1)(-1)^{n+2}\right)=\frac{1}{4}\left(1-[2(n+1)+1](-1)^{n+1}\right)$ ?

$\displaystyle \frac{1}{4}\left(1-(2n+1)(-1)^n\right)+(n+1)(-1)^{n+2}$

$\displaystyle =\frac{1}{4}\left(1-(2n+1)(-1)^n+4(n+1)(-1)^{n+1}(-1)\right)$

$\displaystyle =\frac{1}{4}\left(1+(2n+1)(-1)^{n+1}+4(n+1)(-1)(-1)^{n+1}\right)$

$\displaystyle =\frac{1}{4}\left(1+(-1)^{n+1}(2n+1-4n-4)\right)$

$\displaystyle =\frac{1}{4}\left(1-(2n+3)(-1)^{n+1}\right)$

$\displaystyle =\frac{1}{4}\left(1-[2(n+1)+1](-1)^{n+1}\right)$

True.

Finally prove for n=1.

4. Originally Posted by christina
Prove:

5/(1.2.3) + 6/(2.3.4) +7/(3.4.5) +...+(n+4)/n(n+1)(n+2) = [n(3n+7]/[2(n+1)(n+2)]

Thank you ! =)
$\displaystyle \frac{5}{1(2)3}+\frac{6}{2(3)4}+\frac{7}{3(4)5}+.. .+\frac{n+4}{n(n+1)(n+2)}=\frac{n(3n+7)}{2(n+1)(n+ 2)}$ ?

$\displaystyle \frac{5}{1(2)3}+\frac{6}{2(3)4}+...\frac{n+4}{n(n+ 1)(n+2)}+\frac{n+5}{(n+1)(n+2)(n+3)}=\frac{(n+1)[3(n+1)+7]}{2(n+2)(n+3)}$ ?

$\displaystyle \frac{n(3n+7)}{2((n+1)(n+2)}+\frac{n+5}{(n+1)(n+2) (n+3)}=\frac{(n+1)[3(n+1)+7]}{2(n+2)(n+3)}$ ?

Multiplying all terms by (n+1), multiplying the left fraction by $\displaystyle \frac{n+3}{n+3}$ and the centre fraction by $\displaystyle \frac{2}{2}$

$\displaystyle \frac{n(n+3)(3n+7)}{2(n+2)(n+3)}+\frac{2(n+5)}{2(n +2)(n+3)}=\frac{(n+1)^2[3(n+1)+7]}{2(n+2)(n+3)}$ ?

Now, the numerators need to be checked for equality...

$\displaystyle n(n+3)(3n+7)+2(n+5)=(n+1)^2[3(n+1)+7]$ ?

If you multiply these out, you will find that they both are

$\displaystyle 3n^3+16n^2+23n+10$

True.

now prove for n=1.