Tunjukkan bahwa 1+2+3+...+2n.Tunjukkan bahwa 1+2+3...+n=1/2n (n+1),untuk setiap bilangan asli Bantu jawab ya temen

●langkah dasar
P(1) => anggap benar
n = 1/2n(n+1)
1 = 1/2.1(1+1)
1 = 1/2.1(2)
1 = 1/2.2
1 = 1 -->benar
●langkah induksi
○P(k)=>anggap benar
1+2+3+...+n=1/2l(k+1)
_______Sk=1/2k(k+1)
Sk=1/2k^2 +k
○P(k+1)=>anggap benar
P(k+1)=S(k+1)=1+2+3+...+k=1/2k(k+1)
P(k+1)=S(k+1)=1+2+3+...+k+k+1=1/2k+1(k+1+1)
Sk+k+1 = 1/2 (k+1) (k+2)
1/2 k^2+k+k+k+1=1/2(k+1)(k+2)
1/2 k^2+3k+1 = 1/2 k^2+2k+k+2
k^2+3k+1÷2= k^2+3k+2÷2