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TS EAMCET 4-Aug-2021 Shift 2 Question Paper

Section: Mathematics

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Question : 3 of 160

Marks: +1, -0

For all

n \in \mathbb{N}

2^{2n+1}+3^{2n+1}

is divisible by

7
5
11
8

Solution:

Let

P(n)=2^{2 n+1}+3^{2 n+1}, n \in \mathbb{N}

\;\; =2^{2 n} \cdot 2+3^{2 n} \cdot 3

\;\; =4^{n} \cdot 2+9^{n} \cdot 3

\;\; =2 \cdot (5-1)^{n}+3 \cdot (10-1)^{n}

By using bionomial expansion

(5-1)^{n}={}^{n}C_{0} 5^{n}-{}^{n}C_{1} 5^{n-1}+{}^{n}C_{2} 5^{n-2}+{}^{n}C_{n}(-1)^{n}

(10-1)^{n}={}^{n}C_{0} 10^{n}-{}^{n}C_{1} 10^{n-1}+\dots+{}^{n}C_{n}(-1)^{n}

=2[{}^{n}C_{0} 5^{n}-{}^{n}C_{1} 5^{n-1}+{}^{n}C_{2} 5^{n-2}-{}^{n}C_{3} 5^{n-3}

+\dots+(-1)^{n}]

+3 \cdot [{}^{n}C_{0} 10^{n}-{}^{n}C_{1} 10^{n-1}+{}^{n}C_{2} 10^{n-2}-

{}^{n}C_{3} 10^{n-3}+\dots+(-1)^{n}]

=2 \cdot {}^{n}C_{0} 5^{n}-2 \cdot {}^{n}C_{1} 5^{n-1}+2 \cdot {}^{n}C_{2} 5^{n-2}-2 \cdot {}^{n}C_{3} 5^{n-3}

\dots 2 \cdot (-1)^{n}

+3 \cdot {}^{n}C_{0} 10^{n}-3 \cdot {}^{n}C_{1} 10^{n-1}+3 \cdot {}^{n}C_{2} 10^{2-n}-3 \cdot {}^{n}C_{3} 10^{n-3}

\dots +3 \cdot (-1)^{n}

={}^{n}C_{0}(2 \cdot 5^{n}+3 \cdot 10^{n})-{}^{n}C_{1}(2 \cdot 5^{n-1}+3 \cdot 10^{n-1})

+{}^{n}C_{2}(2 \cdot 5^{n-2}+3 \cdot 10^{n-2})

-{}^{n}C_{3}(2 \cdot 5^{n-3}+3 \cdot 10^{n-3})+\dots+(-1)^{n} 5

=5[{}^{n}C_{0}(2 \cdot 5^{n-1}+3 \cdot 2^{n} \cdot 5^{n-1})

-{}^{n}C_{1}(2 \cdot 5^{n-2}+3 \cdot 2^{n-1} \cdot 5^{n-2})

+{}^{n}C_{2}(2 \cdot 5^{n-3}+3 \cdot 2^{n-2} \cdot 5^{n-3})

P(n)=5 K=-C_{3}(2 \cdot 5^{n-4}+3 \cdot 2^{n-3} \cdot 5^{n-4})+\dots+(-1)^{n}

P(n)=5 K

where,

K={}^{n}C_{0}(2 \cdot 5^{n-1}+3 \cdot 2^{n} 5^{n-1})

-{}^{n}C_{1}(2 \cdot 5^{n-2}+3 \cdot 2^{n-1} \cdot 5^{n-2})+\dots+(-1)^{n}]

\therefore 2^{2 n+1}+3^{3 n+1}

is divisible by

5, \forall n \in \mathbb{N}

.

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