Suppose that result holds for i=k-1 For i=k, we chose internal vertex v so that its children are leaves. we have a new tree which is obtained by omitting edges of incident v and its children By the inductive hypothesis, E =T+2(k-1) We denote the length of path from root to v by l E'=E-l-2,I=I-l. E=E+H2=+2(k-1)+2=l-+2(k-1)+l+2=I+2k Let T be a full m-ary tree. Then E=(m-1)l+mi, where i is the number of internal vertices▪ Suppose that result holds for i=k-1 ▪ For i=k, we chose internal vertex v so that its children are leaves. We have a new tree which is obtained by omitting edges of incident v and its children. ▪ By the inductive hypothesis, E'=I'+2(k-1) ▪ We denote the length of path from root to v by l. ▪ E'=E- l-2, I'=I-l. ▪ E= E'+l+2=I'+2(k-1)+l+2=I-l+2(k-1)+l+2=I+2k。 ▪ Let T be a full m-ary tree. Then E=(m-1)I+mi, where i is the number of internal vertices