Answer to Question #269254 in Discrete Mathematics for Aljon

Question #269254

1. Expand the following using the Binomial Theorem



a. (x-3y)^3



b. (2a+b^-3)^4



c. (a+b+c)²



2. Solve the recurrence relation Pn = Pn-1 + n with the initial condition p_1 = 2 using iteration.

1
Expert's answer
2021-11-22T20:17:55-0500

a.


(x3y)3=x39x2y+27xy227y3(x-3y)^3=x^3-9x^2y+27xy^2-27y^3

b.


(2a+b3)4=16a4+32a3b3+24a2b6+8ab9+b12(2a+b^{-3})^4=16a^4+32a^3b^{-3}+24a^2b^{-6}+8ab^{-9}+b^{-12}

c.


(a+b+c)2=a2+b2+c2+2ab+2ac+2bc(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc

d.


p2=2+2=4p_2=2+2=4

p3=4+3=7p_3=4+3=7

p4=7+4=11p_4=7+4=11

p5=11+5=16p_5=11+5=16

\vdots

(p2p1)+(p3p2)+(p4p3)+...(p_2-p_1)+(p_3-p_2)+(p_4-p_3)+...

+(pn1pn2)+(pnpn1)=2+3+4+...+(p_{n-1}-p_{n-2})+(p_n-p_{n-1})=2+3+4+...

+(n1)+n+(n-1)+n

pnp1=i=1ni1p_n-p_1=\displaystyle\sum_{i=1}^ni-1

pnp1=n(n+1)21p_n-p_1=\dfrac{n(n+1)}{2}-1

pn=1+n(n+1)2p_n=1+\dfrac{n(n+1)}{2}


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