Question

$\left( x+1 \right)$ is a factor of ${{x}^{n}}+1$ only if
(a) n is an odd integer
(b) n is an even integer
(c) n is a negative integer
(d) n is a positive integer.

Answer 1

Hint: Use binomial theorem to prove the algebraic identity
${{a}^{n}}-{{b}^{n}}=\left( a-b \right)\left( {}^{n}{{C}_{1}}{{b}^{n-1}}+.......+{}^{n}{{C}_{n}}{{\left( a-b \right)}^{n-1}} \right)...........(i)$
By using this relate the question to options.

Complete step- by-step solution -
Proof of binomial theorem by Mathematical Induction:
The expression we aim to prove:
${{\left( a+b \right)}^{n}}={{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+.........+{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+........+{}^{n}{{C}_{n-1}}a{{b}^{n-1}}+{{b}^{n}}$
We will first prove that relation true for n = 1, n = 2.
Case 1: n = 1
${{\left( a+b \right)}^{1}}={{a}^{1}}+{{b}^{1}}$
Which is true
as the left-hand side is equal to the right hand side.
Case 2: n = 2
${{\left( a+b \right)}^{2}}={{a}^{2}}+{}^{2}{{C}_{1}}{{a}^{2-1}}b+{{b}^{2}}$
By simplifying, we get:
${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
which is true
it is a general algebraic identity
Now we should take the case where,
n = k,
By substituting this, we get:
${{\left( a+b \right)}^{k}}={{a}^{k}}+{}^{k}{{C}_{1}}{{a}^{k-1}}{{b}^{1}}+{}^{k}{{C}_{2}}{{a}^{k-2}}{{b}^{2}}+{}^{k}{{C}_{k-1}}a{{b}^{k-1}}+{{b}^{k}}$
Now consider the expansion for $n= k+ 1$:
${{\left( a+b \right)}^{k+1}}=\left( a+b \right){{\left( a+b \right)}^{k}}$
By substituting the case of n = k into this equation we get:
${{\left( a+b \right)}^{k}}=\left( a+b \right)\left( {{a}^{k}}+{}^{k}{{C}_{1}}{{a}^{k-1}}b+{}^{k}{{C}_{2}}{{a}^{k-2}}{{b}^{2}}+........+{{b}^{k}} \right)$
By using distributive law, we get:
(a + b).c = ac+ bc
${{\left( a+b \right)}^{k}}={{a}^{k+1}}+\left( 1+{}^{k}{{C}_{1}} \right){{a}^{k}}b+\left( {}^{k}{{C}_{1}}+{}^{k}{{C}_{2}} \right){{a}^{k-1}}{{b}^{2}}+.......+{{b}^{k+1}}$
We know pascal’s identity on combinations is:
${}^{n}{{C}_{r-1}}+{}^{n}{{C}_{r}}={}^{n+1}{{C}_{r}}$
By using Pascal’s identity, we get:
${{\left( a+b \right)}^{k+1}}={{a}^{k+1}}+{}^{k+1}{{C}_{1}}{{a}^{k}}b+........+{}^{k+1}{{C}_{r}}{{a}^{k+1-r}}{{b}^{r}}+..........+{{b}^{k+1}}$
So, by assuming n = k is correct we proved n = k + 1 is correct
So, the binomial theorem is proved by Mathematical Induction.
According to binomial theorem, we have:
${{\left( a+b \right)}^{n}}=\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{a}^{k}}{{b}^{n-k}}}$
We can say:
$a=\left[ \left( a-b \right)+b \right]$
Putting this in left hand side of equation
${{a}^{n}}-{{b}^{n}}=\left( a-b \right)\left[ {}^{n}{{C}_{1}}{{b}^{n-1}}+.............+{}^{n}{{C}_{n}}{{\left( a-b \right)}^{n-1}} \right]$
We get:
${{a}^{n}}-{{b}^{n}}={{\left[ \left( a-b \right)+b \right]}^{n}}-{{b}^{n}}$
By using binomial theorem, we get:
${{a}^{n}}-{{b}^{n}}=\sum{{}^{n}{{C}_{k}}{{\left( a-b \right)}^{k}}{{b}^{n-k}}}$
Putting this left-hand side of equation (i) we get:
${{a}^{n}}-{{b}^{n}}=\left( \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{\left( a-b \right)}^{k}}{{b}^{n-k}}-{{b}^{n}}} \right)$
at k = 0 you can see a term ${{b}^{n}}$ in summation which gets cancelled, and then all the remaining terms are divisible by a – b.
So, by taking (a – b) common from expression we get:
${{a}^{n}}-{{b}^{n}}=\left( a-b \right)\left( {}^{n}{{C}_{1}}{{b}^{n-1}}+........+{}^{n}{{C}_{n}}{{\left( a-b \right)}^{n-1}} \right)$
So, we can say ${{a}^{n}}-{{b}^{n}}$ is divisible by a – b.
If we assume:
$a=x$ , b = -1
By substituting values of a, b in statement we get
${{x}^{n}}-{{\left( -1 \right)}^{n}}$ is divisible by $x-\left( -1 \right)$
By reframing it we get
$x+1$ is a factor of ${{x}^{n}}-{{\left( -1 \right)}^{n}}$
Expression asked in the question:
$x+1$ is a factor of ${{x}^{n}}+1$
So, this is true if n is an odd integer.
Option (a) is correct.

Note: Be careful, while using ${{a}^{n}}-{{b}^{n}}$ .If you don’t observe that ${{b}^{n}}$is cancelled you might not end up proving a – b is factor. Mathematical induction is the easy way to solve this type of problem.