Learn Principle of Mathematical Induction with Free Lessons & Tips

Certainly! Let's prove the given statement using the Principle of Mathematical Induction.

Base Case: For n=1n=1, 1×1!=(1+1)!−1=2!−1=2−1=11×1!=(1+1)!−1=2!−1=2−1=1 So, the base case holds true.

Inductive Step: Assume the given statement is true for some arbitrary natural number kk. That is, assume 1×1!+2×2!+3×3!+⋯+k×k!=(k+1)!−11×1!+2×2!+3×3!+⋯+k×k!=(k+1)!−1

We need to show that the statement holds true for k+1k+1. So, let's consider: 1×1!+2×2!+3×3!+⋯+k×k!+(k+1)×(k+1)!1×1!+2×2!+3×3!+⋯+k×k!+(k+1)×(k+1)!

Now, we can rewrite (k+1)×(k+1)!(k+1)×(k+1)! as (k+1)!(k+1)!. So,

Certainly! Let's solve this using the principle of mathematical induction.

First, we'll prove the base case: when n=1n=1.

We have: 1(1+1)(2⋅1+1)=1⋅2⋅3=61(1+1)(2⋅1+1)=1⋅2⋅3=6

So, n(n+1)(2n+1)n(n+1)(2n+1) is divisible by 6 when n=1n=1.

Now, let's assume that n(n+1)(2n+1)n(n+1)(2n+1) is divisible by 6 for some arbitrary positive integer kk, i.e., k(k+1)(2k+1)k(k+1)(2k+1) is divisible by 6.

Next, we'll prove the inductive step: we'll show that if it holds for kk, it also holds for k+1k+1.

We have: (k+1)((k+1)+1)(2(k+1)+1)(k+1)((k+1)+1)(2(k+1)+1) =(k+1)(k+2)(2k+3)=(k+1)(k+2)(2k+3)

Expanding this expression: =(k+1)(k+2)(2k+3)=(k+1)(k+2)(2k+3) =(k(k+1)(2k+1))+6(k+1)=(k(k+1)(2k+1))+6(k+1)

Since we assumed k(k+1)(2k+1)k(k+1)(2k+1) is divisible by 6, and 6(k+1)6(k+1) is obviously divisible by 6, the entire expression is divisible by 6.

Therefore, by the principle of mathematical induction, n(n+1)(2n+1)n(n+1)(2n+1) is divisible by 6 for all nn belonging to the set of natural numbers.

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As a seasoned tutor registered on UrbanPro, I can confidently affirm that UrbanPro provides top-notch online coaching tuition services. Now, let's delve into the mathematical proof you've presented.

To prove the equation 1+3+32+…+3n−1=3n−121+3+32+…+3n−1=23n−1 using mathematical induction, we follow these steps:

Step 1: Base Case:
First, we verify that the equation holds true for the base case. Let's check for n=1n=1.

1=3(1)−12=22=11=23(1)−1=22=1

The base case holds true.

Step 2: Inductive Hypothesis:
Assume that the equation holds true for some arbitrary positive integer kk. That is,

1+3+32+…+3k−1=3k−121+3+32+…+3k−1=23k−1

Step 3: Inductive Step:
Now, we need to prove that if the equation holds true for kk, then it also holds true for k+1k+1.

Consider the sum 1+3+32+…+3k−1+(3(k+1)−1)1+3+32+…+3k−1+(3(k+1)−1).
By our assumption (inductive hypothesis), this sum is equal to 3k−12+(3(k+1)−1)23k−1+(3(k+1)−1).

Expanding and simplifying, we get:

As an experienced tutor registered on UrbanPro, I can confidently affirm that the number of subsets of a set containing nn distinct elements is indeed 2n2n for all n∈Nn∈N. Let's delve into the proof:

Consider a set SS containing nn distinct elements. To construct a subset of SS, we have two choices for each element: include it in the subset or exclude it. This is true for all nn elements in SS. Thus, for each element, there are 22 possibilities: either it is included in the subset or it is not.

Since there are nn elements in total, by the fundamental principle of counting, the total number of subsets that can be formed from SS is the product of the number of choices for each element, which is 2×2×…×22×2×…×2 (n times), denoted as 2n2n.

Hence, we have proved that the number of subsets of a set containing nn distinct elements is 2n2n for all n∈Nn∈N.

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As an experienced tutor registered on UrbanPro, I can confidently demonstrate that the expression n55+n33+7n155n5+3n3+157n yields a natural number for all nn belonging to the set of natural numbers.

To prove this, let's consider each term of the expression separately:

1. n555n5: Since n5n5 is always a natural number (since nn is a natural number), dividing it by 5 can result in a fraction, but since 5 is a factor of n5n5, the division yields an integer.

2. n333n3: Similar to the previous term, n3n3 is always a natural number, and dividing it by 3 might yield a fraction, but since 3 is a factor of n3n3, the division also yields an integer.

3. 7n15157n: Here, 15 is a multiple of both 5 and 3. So, when we divide 7n7n by 15, it will result in an integer, as both 7n7n and 15 are divisible by 7 and 15 respectively.

Since all three terms of the expression yield integers individually, their sum will also be an integer. And since it's a sum of natural numbers, it will be a natural number.

Therefore, n55+n33+7n155n5+3n3+157n is indeed a natural number for all nn belonging to the set of natural numbers. This demonstrates the validity of the expression for all nn in NN.