Chapter 5, Problem 1RQ

1. Can you use theprinciple of mathematical induction to find a formula for the sum of the firstnterms of a sequence?
2. Can you use the principle of mathematical induction to determine whether a given formula for the sum of the firstnterms of a sequence is correct?c)Find a formul a for the sum of the firstneven positi veintegers, and prove it using mathematical induction,

To determine

(a)

Whether you can use the principal of mathematical induction to find formulae for the sum

of the first n terms of a sequences?

Answer to Problem 1RQ

We cannot use principal of mathematical induction to find a formulae.

Explanation of Solution

Given information:

Sum of the first n terms of a sequences.

The principal of mathematical induction cannot be used to find any formulae.

It can only be used to verify the correctness of the formulae.

Hence, we cannot use principal of mathematical induction to find a formulae for the sum

of the first n terms of a sequences.

To determine

(b)

Whether you can use the principal of mathematical induction to determine a given formulae

for the sum of the first n terms of a sequence is correct?

Answer to Problem 1RQ

We can use principal of mathematical induction to find a given formulae.

Explanation of Solution

Given information:

Sum of the first n terms of a sequences.

The principal of mathematical induction can be used to determine whether

a given formulae.

Hence, we can use principal of mathematical induction to find a given formulae for the sum

of the first n terms of a sequences is correct.

To determine

(c)

A formulae, for the sum of the first n even positive integers and also prove

using mathematical induction.

Answer to Problem 1RQ

Sum of first n even positive integers is true.

Explanation of Solution

Given information:

Sequence is sum of first n even positive integers.

Formulae used:

Mathematical induction.

Calculation:

We know that the sum of n positive integers is,

$1+2+3+\mathrm{.................}+n=\frac{n\left(n+1\right)}{2}$

But even numbers are, $2,4,6,\mathrm{8........},2n$

We can assume that half are the even integers and half are the odd integers in given n

integers.

Therefore, the sum of positive even integers is,

$\begin{array}{l}p\left(n\right)=\frac{2n\times 2\left(n+1\right)}{4}\\ p\left(n\right)=n\left(n+1\right)---------(1)\end{array}$

Now we prove that p(1) is true and the conditional statement

$p(k)\Rightarrow p(k+1)istrueforn=1,2,\mathrm{3.....}$

Basis step is as follows-

P(1) is true from (1) because,

$\begin{array}{l}p(1)=1(1+1)\\ \Rightarrow 2=1(1+1)\end{array}$

Left hand side of this equation is 2 because 2 is the sum of first positive even integers.

Right hand side is found out by substituting 1 for n is n(n+1)

Inductive step is as follows-

Now, let us assume that p(k) holds for an arbitrary positive integer k,

i.e., we assume that

$2+4+6+\mathrm{.....}+2k=k(k+1)-------(2)$

Under this assumption we will show that p(k+1) is true, i.e.,

$\begin{array}{l}2+4+6+\mathrm{.......}+2k+2(k+1)=(k+1)\left[(k+1)+1\right]\\ \Rightarrow 2+4+6+\mathrm{.......}+2k+2(k+1)=(k+1)(k+2)\end{array}$

Is also true.

Now when we add 2(k+1) to both sides of the equation of p(k) in (2),we get

$\begin{array}{l}2+4+6+\mathrm{........2}k+2(k+1)=k(k+1)+2k+2\\ \Rightarrow 2+4+6+\mathrm{........2}k+2(k+1)=(k+1)(k+2)------(3)\end{array}$

The equation (3) shows that p(k+1) is true under the assumption that p(k) is true.

this complete inductive step.

Hence, we completed both basis and inductive steps, so by mathematical induction we

Know that p(n) is true for all even positive integers n.