Logo

AskSia

Plus

solve a c e g ample 18 1 Prove each of the following using mathematical inductio...
Jul 1, 2024
solve a c e g
Solution by Steps
step 1
Base Case: For n=1n=1, 1=1(1+1)2=11 = \frac{1(1+1)}{2} = 1. The base case holds
step 2
Inductive Step: Assume the formula holds for n=kn=k, i.e., 1+2+3+⋯+k=k(k+1)21+2+3+\cdots+k=\frac{k(k+1)}{2}
step 3
We need to prove it for n=k+1n=k+1: 1+2+3+⋯+k+(k+1)=(k+1)(k+2)21+2+3+\cdots+k+(k+1)=\frac{(k+1)(k+2)}{2}
step 4
Using the inductive hypothesis: 1+2+3+⋯+k+(k+1)=k(k+1)2+(k+1)1+2+3+\cdots+k+(k+1)=\frac{k(k+1)}{2}+(k+1)
step 5
Simplify: k(k+1)2+(k+1)=k(k+1)+2(k+1)2=(k+1)(k+2)2\frac{k(k+1)}{2}+(k+1)=\frac{k(k+1)+2(k+1)}{2}=\frac{(k+1)(k+2)}{2}
step 6
Therefore, the formula holds for n=k+1n=k+1. By induction, the formula is true for all nn
Answer
The formula 1+2+3+⋯+n=n(n+1)21+2+3+\cdots+n=\frac{n(n+1)}{2} is proven by induction.
Key Concept
Mathematical Induction
Explanation
Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers.
Question c: Prove 12+22+32+⋯+n2=n(n+1)(2n+1)61^{2}+2^{2}+3^{2}+\cdots+n^{2}=\frac{n(n+1)(2n+1)}{6} by induction
step 1
Base Case: For n=1n=1, 12=1(1+1)(2⋅1+1)6=11^2 = \frac{1(1+1)(2\cdot1+1)}{6} = 1. The base case holds
step 2
Inductive Step: Assume the formula holds for n=kn=k, i.e., 12+22+32+⋯+k2=k(k+1)(2k+1)61^2+2^2+3^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}
step 3
We need to prove it for n=k+1n=k+1: 12+22+32+⋯+k2+(k+1)2=(k+1)(k+2)(2(k+1)+1)61^2+2^2+3^2+\cdots+k^2+(k+1)^2=\frac{(k+1)(k+2)(2(k+1)+1)}{6}
step 4
Using the inductive hypothesis: 12+22+32+⋯+k2+(k+1)2=k(k+1)(2k+1)6+(k+1)21^2+2^2+3^2+\cdots+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2
step 5
Simplify: k(k+1)(2k+1)6+(k+1)2=k(k+1)(2k+1)+6(k+1)26=(k+1)(k(2k+1)+6(k+1))6\frac{k(k+1)(2k+1)}{6}+(k+1)^2=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}=\frac{(k+1)(k(2k+1)+6(k+1))}{6}
step 6
Further simplify: (k+1)(2k2+k+6k+6)6=(k+1)(2k2+7k+6)6=(k+1)(k+2)(2k+3)6\frac{(k+1)(2k^2+k+6k+6)}{6}=\frac{(k+1)(2k^2+7k+6)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}
step 7
Therefore, the formula holds for n=k+1n=k+1. By induction, the formula is true for all nn
Answer
The formula 12+22+32+⋯+n2=n(n+1)(2n+1)61^{2}+2^{2}+3^{2}+\cdots+n^{2}=\frac{n(n+1)(2n+1)}{6} is proven by induction.
Key Concept
Mathematical Induction
Explanation
Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers.
Question e: Prove 1⋅2+2⋅3+⋯+n(n+1)=n(n+1)(n+2)31 \cdot 2+2 \cdot 3+\cdots+n(n+1)=\frac{n(n+1)(n+2)}{3} by induction
step 1
Base Case: For n=1n=1, 1⋅2=1(1+1)(1+2)3=21 \cdot 2 = \frac{1(1+1)(1+2)}{3} = 2. The base case holds
step 2
Inductive Step: Assume the formula holds for n=kn=k, i.e., 1⋅2+2⋅3+⋯+k(k+1)=k(k+1)(k+2)31 \cdot 2+2 \cdot 3+\cdots+k(k+1)=\frac{k(k+1)(k+2)}{3}
step 3
We need to prove it for n=k+1n=k+1: 1⋅2+2⋅3+⋯+k(k+1)+(k+1)(k+2)=(k+1)(k+2)(k+3)31 \cdot 2+2 \cdot 3+\cdots+k(k+1)+(k+1)(k+2)=\frac{(k+1)(k+2)(k+3)}{3}
step 4
Using the inductive hypothesis: 1⋅2+2⋅3+⋯+k(k+1)+(k+1)(k+2)=k(k+1)(k+2)3+(k+1)(k+2)1 \cdot 2+2 \cdot 3+\cdots+k(k+1)+(k+1)(k+2)=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)
step 5
Simplify: k(k+1)(k+2)3+(k+1)(k+2)=k(k+1)(k+2)+3(k+1)(k+2)3=(k+1)(k+2)(k+3)3\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)=\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}=\frac{(k+1)(k+2)(k+3)}{3}
step 6
Therefore, the formula holds for n=k+1n=k+1. By induction, the formula is true for all nn
Answer
The formula 1⋅2+2⋅3+⋯+n(n+1)=n(n+1)(n+2)31 \cdot 2+2 \cdot 3+\cdots+n(n+1)=\frac{n(n+1)(n+2)}{3} is proven by induction.
Key Concept
Mathematical Induction
Explanation
Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers.
Question g: Prove 11⋅3+13⋅5+⋯+1(2n−1)(2n+1)=n2n+1\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\cdots+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1} by induction
step 1
Base Case: For n=1n=1, 11⋅3=13=12⋅1+1\frac{1}{1 \cdot 3} = \frac{1}{3} = \frac{1}{2 \cdot 1 + 1}. The base case holds
step 2
Inductive Step: Assume the formula holds for n=kn=k, i.e., 11⋅3+13⋅5+⋯+1(2k−1)(2k+1)=k2k+1\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\cdots+\frac{1}{(2k-1)(2k+1)}=\frac{k}{2k+1}
step 3
We need to prove it for n=k+1n=k+1: 11⋅3+13⋅5+⋯+1(2k−1)(2k+1)+1(2(k+1)−1)(2(k+1)+1)=k+12(k+1)+1\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\cdots+\frac{1}{(2k-1)(2k+1)}+\frac{1}{(2(k+1)-1)(2(k+1)+1)}=\frac{k+1}{2(k+1)+1}
step 4
Using the inductive hypothesis: 11⋅3+13⋅5+⋯+1(2k−1)(2k+1)+1(2k+1)(2k+3)=k2k+1+1(2k+1)(2k+3)\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\cdots+\frac{1}{(2k-1)(2k+1)}+\frac{1}{(2k+1)(2k+3)}=\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}
step 5
Simplify: k2k+1+1(2k+1)(2k+3)=k(2k+3)+1(2k+1)(2k+3)=2k2+3k+1(2k+1)(2k+3)\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}=\frac{k(2k+3)+1}{(2k+1)(2k+3)}=\frac{2k^2+3k+1}{(2k+1)(2k+3)}
step 6
Further simplify: 2k2+3k+1(2k+1)(2k+3)=(k+1)(2k+1)(2k+1)(2k+3)=k+12k+3\frac{2k^2+3k+1}{(2k+1)(2k+3)}=\frac{(k+1)(2k+1)}{(2k+1)(2k+3)}=\frac{k+1}{2k+3}
step 7
Therefore, the formula holds for n=k+1n=k+1. By induction, the formula is true for all nn
Answer
The formula 11⋅3+13⋅5+⋯+1(2n−1)(2n+1)=n2n+1\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\cdots+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1} is proven by induction.
Key Concept
Mathematical Induction
Explanation
Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers.
© 2023 AskSia.AI all rights reserved