- For I 2 I N I 3 Assume N 2 3k For J 1 J I J 5 Statements Answer N 2 3k K Log3 2n F N 5 2 30 2 31 1 (25.76 KiB) Viewed 29 times
for (i=2;i⇔=n;i∗=3){// Assume n=2×3k for (j=1;j<=i;j++){ // 5 statements \} Answer: n=2×3k→k=log32nf(n)=5[2×30+2×31+⋯
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for (i=2;i⇔=n;i∗=3){// Assume n=2×3k for (j=1;j<=i;j++){ // 5 statements \} Answer: n=2×3k→k=log32nf(n)=5[2×30+2×31+⋯
for (i=2;i⇔=n;i∗=3){// Assume n=2×3k for (j=1;j<=i;j++){ // 5 statements \} Answer: n=2×3k→k=log32nf(n)=5[2×30+2×31+⋯+2×3k]=5[3×3log32n−1]=5(3×2n−1)
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