Dear Student, Given, an=2n-1an-1⇒log2an=log22n-1an-1⇒log2an=log22n-1+log2an-1⇒log2an=n-1+log2an-1⇒log2an-1=log2an-n+1So, log2an-2=log2an-1-n+1⇒log2an-2=log2an-n+1-n+1⇒log2an-2=log2an-2 (n-1)Similarly, log2an-3=log2an-2-n+1⇒log2an-3=log2an-3 (n-1) log2an-(n-1)=log2an-(n-1) (n-1)⇒ log2a1=log2an-(n-1) (n-1)Given a1=1 , So, log2a1=0⇒ 0=log2an-(n-1) (n-1)⇒log2an=(n-1) (n-1)So, log2a100=(100-1) (100-1)=9801 Regards,

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Dear Student,

G i v e n, a_{n} = 2^{n - 1} a_{n - 1} \Rightarrow \log_{2} (a_{n}) = \log_{2} (2^{n - 1} a_{n - 1}) \Rightarrow \log_{2} (a_{n}) = \log_{2} (2^{n - 1}) + \log_{2} (a_{n - 1}) \Rightarrow \log_{2} (a_{n}) = n - 1 + \log_{2} (a_{n - 1}) \Rightarrow \log_{2} (a_{n - 1}) = \log_{2} (a_{n}) - n + 1 S o, \log_{2} (a_{n - 2}) = \log_{2} (a_{n - 1}) - n + 1 \Rightarrow \log_{2} (a_{n - 2}) = \log_{2} (a_{n}) - n + 1 - n + 1 \Rightarrow \log_{2} (a_{n - 2}) = \log_{2} (a_{n}) - 2 . (n - 1) S i m i l a r l y, \log_{2} (a_{n - 3}) = \log_{2} (a_{n - 2}) - n + 1 \Rightarrow \log_{2} (a_{n - 3}) = \log_{2} (a_{n}) - 3 . (n - 1) . . . . \log_{2} (a_{n - (n - 1)}) = \log_{2} (a_{n}) - (n - 1) . (n - 1) \Rightarrow \log_{2} (a_{1}) = \log_{2} (a_{n}) - (n - 1) . (n - 1) G i v e n a_{1} = 1, S o, \log_{2} (a_{1}) = 0 \Rightarrow 0 = \log_{2} (a_{n}) - (n - 1) . (n - 1) \Rightarrow \log_{2} (a_{n}) = (n - 1) . (n - 1) S o, \log_{2} (a_{100}) = (100 - 1) . (100 - 1) = 9801

Regards,

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