Question

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Question No. 4 (i)

Needs a Explained Answer ! ​

Answer 1

Fibonacci Sequence

    Fibonacci Sequence is a sequence such that, on considering any three consecutive terms from the sequence, the sum of the first two terms is equal to the third term. It looks like,  

1, 1, 2, 3, 5, 8, 13, 21,...

The terms are taken here as  $a_1,\ a_2,\ a_3,\ a_4,...$

So, on taking any one of the term from the sequence as  $a_n,$  

$a_n=a_{n-1}+a_{n-2}$

where  $n\geq 3$

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The given identities can be proved by principle of mathematical induction.  

$(i)\ \ a_1+a_2+a_3+...+a_n=a_{n+2}-1$

$\textsf{Let}\ \ n=1. \\ \\ a_1=\mathbf{1}\\ \\ a_3-1=2-1=\mathbf{1}\\ \\ \\ \textsf{Let}\ \ n=k.\\ \\ \textsf{Assume that}\ \ a_1+a_2+a_3+...+a_k=a_{k+2}-1\\ \\ \\ \textsf{Let}\ \ n=k+1. \\ \\ \begin{aligned}&a_1+a_2+a_3+...+a_{k+1}\\ \\ \Rightarrow\ \ &a_1+a_2+a_3+...+a_k+a_{k+1}\\ \\ \Rightarrow\ \ &a_{k+2}-1+a_{k+1}\\ \\ \Rightarrow\ \ &a_{k+1}+a_{k+2}-1\\ \\ \Rightarrow\ \ &a_{k+3}-1\ \ \ \ \ [\because\ a_{n-2}+a_{n-1}=a_n]\\ \\ \Rightarrow\ \ &a_{n+2}-1\end{aligned}$

$(ii)\ \ a_1+a_3+a_5+...+a_{2n-1}=a_{2n}$

$(ii)\\ \\ \textsf{Let}\ \ n=1.\\ \\ a_1=\mathbf{1}\\ \\ a_{2n}=a_2=\mathbf{1}\\ \\ \\ \textsf{Let}\ \ n=k.\\ \\ \textsf{Assume that}\ \ a_1+a_3+a_5+...+a_{2k-1}=a_{2k}\\ \\ \\ \textsf{Let}\ \ n=k+1.\\ \\ \begin{aligned}&a_1+a_3+a_5+...+a_{2(k+1)-1}\\ \\ \Rightarrow\ \ &a_1+a_3+a_5+...+a_{2k-1}+a_{2k+1}\\ \\ \Rightarrow\ \ &a_{2k}+a_{2k+1}\\ \\ \Rightarrow\ \ &a_{2k+2}\ \ \ \ \ [\because\ a_{n-2}+a_{n-1}=a_n]\\ \\ \Rightarrow\ \ &a_{2(k+1)}\\ \\ \Rightarrow\ \ &a_{2n}\end{aligned}$

Also,  

$(iii)\ \ a_2+a_4+a_6+...+a_{2n}=a_{2n+1}-1$

$\textsf{Let}\ \ n=1.\\ \\ a_2=\mathbf{1}\\ \\ a_3-1=2-1=\mathbf{1}\\ \\ \\ \textsf{Let}\ \ n=k.\\ \\ \textsf{Assume that}\ \ a_2+a_4+a_6+...+a_{2k}=a_{2k+1}-1\\ \\ \\ \textsf{Let}\ \ n=k+1.\\ \\ \begin{aligned}&a_2+a_4+a_6+...+a_{2(k+1)}\\ \\ \Rightarrow\ \ &a_2+a_4+a_6+...+a_{2k}+a_{2k+2}\\ \\ \Rightarrow\ \ &a_{2k+1}-1+a_{2k+2}\\ \\ \Rightarrow\ \ &a_{2k+1}+a_{2k+2}-1\\ \\ \Rightarrow\ \ &a_{2k+3}-1\\ \\ \Rightarrow\ \ &a_{2(k+1)+1}-1\\ \\ \Rightarrow\ \ &a_{2n+1}-1\end{aligned}$

$(iv)\ \ (a_{n+1})^2-(a_n\cdot a_{n+2})=(-1)^n$

$\textsf{Let}\ \ n=1.\\ \\ (a_2)^2-(a_1\cdot a_3)=1^2-1\cdot 2=1-2=\mathbf{-1=(-1)^1}\\ \\ \\ \textsf{Let}\ \ n=k.\\ \\ \textsf{Assume that}\ \ (a_{k+1})^2-(a_k\cdot a_{k+2})=(-1)^k\\ \\ \\ \textsf{Let}\ \ n=k+1.$

$\begin{aligned}&(a_{k+2})^2-(a_{k+1}\cdot a_{k+3})\\ \\ \Rightarrow\ \ &(a_k+a_{k+1})^2-(a_{k+1}\cdot a_{k+3})\\ \\ \Rightarrow\ \ &(a_k)^2+2a_k\cdot a_{k+1}+(a_{k+1})^2-(a_{k+1}\cdot a_{k+3})\\ \\ \Rightarrow\ \ &(a_k)^2+a_{k+1}(2a_k+a_{k+1}-a_{k+3})\\ \\ \Rightarrow\ \ &(a_k)^2+a_{k+1}(a_k+a_k+a_{k+1}-a_{k+3})\\ \\ \Rightarrow\ \ &(a_k)^2+a_{k+1}(a_k+a_{k+2}-a_{k+3})\\ \\ \Rightarrow\ \ &(a_k)^2+a_k\cdot a_{k+1}+a_{k+1}\cdot a_{k+2}-a_{k+1}\cdot a_{k+3}\end{aligned}$

$\begin{aligned}\Rightarrow\ \ &a_k(a_k+a_{k+1})+a_{k+1}(a_{k+2}-a_{k+3})\\ \\ \Rightarrow\ \ &a_k\cdot a_{k+2}-a_{k+1}(a_{k+3}-a_{k+2})\\ \\ \Rightarrow\ \ &a_k\cdot a_{k+2}-a_{k+1}\cdot a_{k+1}\ \ \ \ \ [\because\ a_n-a_{n-1}=a_{n-2}]\\ \\ \Rightarrow\ \ &a_k\cdot a_{k+2}-(a_{k+1})^2\\ \\ \Rightarrow\ \ &-((a_{k+1})^2-(a_k\cdot a_{k+2}))\\ \\ \Rightarrow\ \ &-(-1)^k\\ \\ \Rightarrow\ \ &(-1)^k\cdot (-1)^1\\ \\ \Rightarrow\ \ &(-1)^{k+1}\\ \\ \Rightarrow\ \ &(-1)^n\end{aligned}$

Hence Proved!