Question

Задание на 100 баллов!Помогите,с развернутым решением!

Answer 1

$1) a_n=a_1+(n-1)d, nin N; \ n=1, a_1=a_1+(1-1)d=a_1+0cdot d=a_1+0=a_1; \ n=k, a_k=a_1+(k-1)d, \ n=k+1, a_{k+1}=a_1+((k+1)-1)d=a_1+(k+1-1)d=\=a_1+((k-1)+1)d=a_1+(k-1)d+d=a_k+d, \ a_{k+1}=a_k+d;$

$2) A(n)=3^{n+2}+2^{3n}, \ A(1)=3^{1+2}+2^{3cdot1}=3^{3}+2^{3}=27+8=35vdots5; \ A(k)=(3^{k+2}+2^{3k})vdots5; \ A(k+1)=3^{(k+1)+2}+2^{3(k+1)}=3^{(k+2)+1}+2^{3k+3}=3cdot3^{k+2}+8cdot2^{3k}=\=3cdot3^{k+2}+3cdot2^{3k}+5cdot2^{3k}=3cdot3^{k+2}+2^{3k}+5cdot2^{3k}=A(k)+5cdot2^{3k}, \ A(k)vdots5, 5cdot2^{3k}vdots5Rightarrow A(k+1)vdots5;$

$3) A(n)=2cdot3+3cdot4+dots+(n+1)(n+2)=frac{n(n^2+6n+11)}{3}, \ A(1)=(1+1)(1+2)=frac{1cdot(1^2+6cdot1+11)}{3}, \ 2cdot3=frac{1+6+11}{3}, \ 6=6; \ A(k)=2cdot3+3cdot4+dots+(k+1)(k+2)=frac{k(k^2+6k+11)}{3},$
$A(k+1)=2cdot3+3cdot4+dots+(k+1)(k+2)+((k+1)+1)((k+1)+2)=\=frac{(k+1)((k+1)^2+6(k+1)+11)}{3}, \ frac{k(k^2+6k+11)}{3}+(k+2)(k+3)=frac{(k+1)(k^2+2k+1+6k+6+11)}{3}, \ k(k^2+6k+11)+3(k+2)(k+3)-(k+1)(k^2+8k+18)=0, \ k(k^2+6k+11-k^2-8k-18)+3k^2+15k+18-k^2-8k-18=0, \ k(-2k-7)+2k^2+7k=0, \ -2k^2-7k+2k^2+7k=0, \ 0=0.$