CAT 2019 Set-2 | Question: 80

Question

Let $a_{1},a_{2}\dots$ be integers such that $a_{1}-a_{2}+a_{3}-a_{4}+\dots+(-1)^{n-1}a_{n}=n$, for all $n\geq 1$ Then $a_{51}+a_{52}+\dots+a_{1023}$ equals

• A. $-1$
• B. $10$
• C. $0$
• D. $1$

Answer 1

Given that , $a_{1} – a_{2}+a_{3}- a_{4}+ \dots + (-1)^{n-1} a_{n}=n \quad \longrightarrow (1)$

Now, put the various of $n$ and observe.

• $ n=1 \Rightarrow a_{1} = 1$
• $ n=2 \Rightarrow a_{1} - a_{2} =2 \Rightarrow a_{2}=-1$
• $ n=3 \Rightarrow a_{1} -a_{2} +a_{3} =3 \Rightarrow a_{3} =1$
• $ n=4 \Rightarrow a_{1} -a_{2} +a_{3} -a_{4} =4 \Rightarrow a_{4} = -1$
• $ n=5 \Rightarrow a_{1} – a_{2} +a_{3} - a_{4} +a_{5} =5 \Rightarrow a_{5} =1$
• $ n=6 \Rightarrow a_{1}-a_{2} +a_{3}-a_{4} +a_{5} -a_{6} =6 \Rightarrow a_{6} = -1$ 
• $ \vdots \quad \vdots \quad \vdots $

We observe that , when 

• $n=\text{odd} \Rightarrow a_{\text{odd}} =1$
• $ n=\text{even} \Rightarrow a_{\text{even}} = -1$

Now, $ a_{51}+a_{52} +a_{53} +\dots +a_{1022} +a_{1023}$

$\qquad = \underbrace{1-1+1-1+ \dots + (-1)}_{\text{Sum  of all terms = 0}}+1$

$\qquad = 0 + 1 = 1$

Correct Answer: D