Question

For \[n\in N,{{x}^{n+1}}+{{\left( x+1 \right)}^{2n-1}}\] is divisible by
\[\left( a \right)x\]
\[\left( b \right)x+1\]
\[\left( c \right){{x}^{2}}+x+1\]
\[\left( d \right){{x}^{2}}-x+1\]

Answer 1

Hint: To solve this question first put n = 1 in \[{{x}^{n+1}}+{{\left( x+1 \right)}^{2n-1}}\] and then see which of the following given option satisfy the required condition. Then again proceed for using n = 2 in \[{{\left( x \right)}^{n+1}}+{{\left( x+1 \right)}^{2n-1}}\] and solving further try to obtain the same value as obtained in n = 1, and use the formula \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}\] to expand.

Complete step by step answer:
We are given \[{{x}^{n+1}}+{{\left( x+1 \right)}^{2n-1}}\] for \[n\in N.\] Because \[n\in N,\] the set of natural numbers so let us first use n = 1, n = 2, etc to get our result when n = 1, then we have, \[{{x}^{n+1}}+{{\left( x+1 \right)}^{2n-1}}\] becomes
\[={{x}^{2}}+{{\left( x+1 \right)}^{2\times 1-1}}\]
\[={{x}^{2}}+{{\left( x+1 \right)}^{1}}\]
\[={{x}^{2}}+x+1\]
which is divisible by \[{{x}^{2}}+x+1.\]
Because every number is divisible by itself.
Similarly, let us consider n = 2. Putting n = 2 in \[{{\left( x \right)}^{n+1}}+{{\left( x+1 \right)}^{2n-1}}\] we get,
\[={{x}^{3}}+{{\left( x+1 \right)}^{2\times 2-1}}\]
\[={{x}^{3}}+{{\left( x+1 \right)}^{4-1}}\]
\[={{x}^{3}}+{{\left( x+1 \right)}^{3}}\]
Opening the bracket using the formula of \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}},\] we get,
\[={{x}^{3}}+{{x}^{3}}+1+3{{x}^{2}}+3x\]
\[=2{{x}^{3}}+3{{x}^{2}}+3x+1\]
Taking 2x common and writing 3x as 2x + x, we get,
\[\Rightarrow 2{{x}^{3}}+3{{x}^{2}}+3x+1=2{{x}^{3}}+3{{x}^{2}}+2x+x+1\]
\[\Rightarrow 2{{x}^{3}}+3{{x}^{2}}+3x+1=2x\left( {{x}^{2}}+1 \right)+3{{x}^{2}}+x+1\]
Again writing \[3{{x}^{2}}=2{{x}^{2}}+x\] we get,
\[=2x\left( {{x}^{2}}+1 \right)+2{{x}^{2}}+{{x}^{2}}+x+1\]
\[=2x\left[ {{x}^{2}}+1+x \right]+\left( {{x}^{2}}+x+1 \right)\]
which is nothing but
\[=2x\left( 1+x+{{x}^{2}} \right)+1\left( 1+x+{{x}^{2}} \right)\]
\[\Rightarrow \left( 2x+1 \right)\left( 1+x+{{x}^{2}} \right)\]
So, for n = 2, the term \[{{x}^{n+1}}+{{\left( x+1 \right)}^{2n-1}}\] is divisible by \[\left( 1+x+{{x}^{2}} \right).\] Therefore, as for two terms as n = 1 and n = 2, both have \[{{x}^{n+1}}+{{\left( x+1 \right)}^{2n-1}}\] which is divisible by \[{{x}^{2}}+x+1\] then it is true for all \[n\in N,\] set of natural numbers.
Therefore, for \[n\in N,\] we have \[{{x}^{n+1}}+{{\left( x+1 \right)}^{2n-1}}\] is divisible by \[{{x}^{2}}+x+1.\]

So, the correct answer is “Option C”.

Note: Even if you try to go for n = 3, then \[{{\left( x+1 \right)}^{2n-1}}\] is always of odd power, so 2n – 1 is always odd. Hence we can always sum \[{{\left( a+b \right)}^{3}}\] formula and solve further as for n = 2, then it would be correct \[\forall n\in N.\] Also, \[{{x}^{n+1}}\] is always going to work as \[{{x}^{2}}\] can be taken common always from \[{{x}^{n+1}}.\] Also note that for these types of questions, always go for putting n = 1, 2, 3 and checking if \[n\in W,\] whole numbers, then we would have started from n = 0, 1, 2, …. etc.