Respuesta :
google translate - indonesian:
"Given the original numbers design a formula to find the 400th term and continue (prove) with mathematical induction"
Let [tex]u_n[/tex] represent the n'th term of the sequence, that is
[tex]u_1[/tex] is the first term, [tex]u_2[/tex] the second term and so on...
notice that :
[tex]u_1=3[/tex]
[tex]u_2=3+2[/tex]
[tex]u_3=3+2+3[/tex]
[tex]u_4=3+(2+3+4)[/tex]
[tex]u_5=3+(2+3+4+5)[/tex]
so clearly
[tex]u_n=3+(2+3+4+5...+n)=2+(1+2+3+...+n)[/tex]
Recall the famous Gauss formula for addition of the first consecutive n natural numbers: 1+2+3+...(n-1)+n=n(n+1)/2
then, the formula for u_n is:
[tex]u_n=2+(1+2+3+...+n)=2+ \frac{n(n+1)}{2} [/tex]
for example, the 5th term is :
[tex]u_5=2+ \frac{5(5+1)}{2}=2+15=17[/tex]
Proof by induction.
step 1
[tex]u_1=2+ \frac{1(1+1)}{2}=2+1=3[/tex], true
step2
assume true for n=k, that is assume [tex]u_k=2+ \frac{k(k+1)}{2}[/tex]
step 3
verify for n=k+1, that is verify that [tex]u_k_+_1=2+ \frac{(k+1)(k+2)}{2} [/tex].
[tex]u_k_+_1=u_k+k+1}[/tex] because each nth term of the sequence is clearly its previous term +n
for example the fourth term 12, is the third term +4.
the fifth term, 17, is the fourth term 12 + 5 and so on...
also, by our assumption [tex]u_k=2+ \frac{k(k+1)}{2}[/tex]
so:
[tex]u_k_+_1=u_k+k+1=2+ \frac{k(1+1)}{2}+k+1=2+\frac{k(1+1)}{2}+ \frac{2(k+1)}{2}[/tex]
factorizing k+1
[tex]=2+ \frac{(k+1)((k+2)}{2} [/tex]
which is what we needed to show.
Answer: [tex]u_n=2+ \frac{n(n+1)}{2} [/tex]
"Given the original numbers design a formula to find the 400th term and continue (prove) with mathematical induction"
Let [tex]u_n[/tex] represent the n'th term of the sequence, that is
[tex]u_1[/tex] is the first term, [tex]u_2[/tex] the second term and so on...
notice that :
[tex]u_1=3[/tex]
[tex]u_2=3+2[/tex]
[tex]u_3=3+2+3[/tex]
[tex]u_4=3+(2+3+4)[/tex]
[tex]u_5=3+(2+3+4+5)[/tex]
so clearly
[tex]u_n=3+(2+3+4+5...+n)=2+(1+2+3+...+n)[/tex]
Recall the famous Gauss formula for addition of the first consecutive n natural numbers: 1+2+3+...(n-1)+n=n(n+1)/2
then, the formula for u_n is:
[tex]u_n=2+(1+2+3+...+n)=2+ \frac{n(n+1)}{2} [/tex]
for example, the 5th term is :
[tex]u_5=2+ \frac{5(5+1)}{2}=2+15=17[/tex]
Proof by induction.
step 1
[tex]u_1=2+ \frac{1(1+1)}{2}=2+1=3[/tex], true
step2
assume true for n=k, that is assume [tex]u_k=2+ \frac{k(k+1)}{2}[/tex]
step 3
verify for n=k+1, that is verify that [tex]u_k_+_1=2+ \frac{(k+1)(k+2)}{2} [/tex].
[tex]u_k_+_1=u_k+k+1}[/tex] because each nth term of the sequence is clearly its previous term +n
for example the fourth term 12, is the third term +4.
the fifth term, 17, is the fourth term 12 + 5 and so on...
also, by our assumption [tex]u_k=2+ \frac{k(k+1)}{2}[/tex]
so:
[tex]u_k_+_1=u_k+k+1=2+ \frac{k(1+1)}{2}+k+1=2+\frac{k(1+1)}{2}+ \frac{2(k+1)}{2}[/tex]
factorizing k+1
[tex]=2+ \frac{(k+1)((k+2)}{2} [/tex]
which is what we needed to show.
Answer: [tex]u_n=2+ \frac{n(n+1)}{2} [/tex]
