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Question: The statement \(P\left( n \right)={{9}^{n}}-{{8}^{n}}\) when divided by 8 always leaves the remainde...

The statement P(n)=9n8nP\left( n \right)={{9}^{n}}-{{8}^{n}} when divided by 8 always leaves the remainder.
[a] 2
[b] 3
[c] 1
[d] 7

Explanation

Solution

Hint: Write 9 as 8+1 and use the fact that the expansion of (x+y)n{{\left( x+y \right)}^{n}} is given by (x+y)n=r=0nnCrxnryr{{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}}

Complete step-by-step answer:
Hence write the expansion of (1+8)n{{\left( 1+8 \right)}^{n}}.
Subtract 8n{{8}^{n}} from both sides of the expansion and hence find the remainder obtained on dividing 9n8n{{9}^{n}}-{{8}^{n}} by 8.
Alternatively, put 8 = x and hence write P(n) in terms of x, i.e. g(x)=P(n)=(x+1)nxng\left( x \right)=P\left( n \right)={{\left( x+1 \right)}^{n}}-{{x}^{n}}. Use remainder theorem, which states that the remainder obtained on dividing p(x) by x-a is given by p(a)p\left( a \right). Hence find the remainder obtained on dividing g(x) by x and hence find the remainder obtained on dividing P(n) by 8.

We have P(n)=9n8nP\left( n \right)={{9}^{n}}-{{8}^{n}}
We know that 9 = 1+8
Hence, we have
P(n)=(1+8)n8nP\left( n \right)={{\left( 1+8 \right)}^{n}}-{{8}^{n}}
We know from the binomial theorem that the expansion of (x+y)n{{\left( x+y \right)}^{n}} is given by (x+y)n=r=0nnCrxnryr{{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}}
Hence, we have
P(n)=1+nC18+nC282++nCn18n1+nCn8n8nP\left( n \right)=1{{+}^{n}}{{C}_{1}}8{{+}^{n}}{{C}_{2}}{{8}^{2}}+\cdots {{+}^{n}}{{C}_{n-1}}{{8}^{n-1}}{{+}^{n}}{{C}_{n}}{{8}^{n}}-{{8}^{n}}
Hence, we have
P(n)=8(nC1+nC28++nCn18n2)+1=8k+1,kNP\left( n \right)=8\left( ^{n}{{C}_{1}}{{+}^{n}}{{C}_{2}}8+\cdots {{+}^{n}}{{C}_{n-1}}{{8}^{n-2}} \right)+1=8k+1,k\in \mathbb{N}
Hence, by Euclid's division lemma, the remainder obtained on dividing P(n) by 8 is 1.
Hence option [c] is correct.

Note: Alternative Solution:
Let x = 8
Hence, we have P(n)=g(x)=(x+1)nxnP\left( n \right)=g\left( x \right)={{\left( x+1 \right)}^{n}}-{{x}^{n}}
We know from the remainder theorem that the remainder obtained on dividing p(x) by x-a is given by p(a)p\left( a \right)
Hence, the remainder obtained on dividing g(x) by x is given by g(0)
Now, we have
g(0)=(0+1)n0n=1g\left( 0 \right)={{\left( 0+1 \right)}^{n}}-{{0}^{n}}=1
Hence the remainder obtained on dividing P(n) by 8 is 1.
Hence option [c] is correct.