By-murging-two-sequences-a-1-a-2-a-n-and-b-1-b-2-b-n-a-new-sequence-a-1-b-1-a-2-b-2-a-n-b-n-has-been-produced-Determine-a-single-formula-general-term-of-th

Question Number 4485 by Rasheed Soomro last updated on 31/Jan/16

B y m u r g i n g t w o s e q u e n c e s

a_{1}, a_{2}, \dots a_{n} a n d b_{1}, b_{2}, \dots, b_{n}

a n e w s e q u e n c e

a_{1}, b_{1}, a_{2}, b_{2}, \dots, a_{n}, b_{n} h a s b e e n

p r o d u c e d .

D e t e r m i n e a single - formula

general term o f t h i s n e w s e q u e n c e

i n t e r m s o f a_{n} a n d b_{n} .

Commented by prakash jain last updated on 31/Jan/16

c_{n} = \frac{1}{2} [a_{n} + {(- 1)}^{n + 1} a_{n} + b_{n} + {(- 1)}^{n} b_{n}]

Commented by FilupSmith last updated on 01/Feb/16

How did you come to this solution ?

Commented by prakash jain last updated on 01/Feb/16

My solution is wrong . I gave solution for

it was a_{1}, b_{2}, a_{3}, c_{4}

Commented by Yozzii last updated on 01/Feb/16

c_{r} = a_{⌊ \frac{r + 1}{2} ⌋} ∣ s i n (\frac{r π}{2}) ∣ + b_{⌊ \frac{r + 1}{2} ⌋} ∣ s i n (\frac{(r + 1) π}{2}) ∣

r = 1, 2, 3, 4, \dots, 2 n - 3, 2 n - 2, 2 n - 1, 2 n .

c_{1} = a_{1} \times 1 + b_{1} \times 0 = a_{1}

c_{2} = a_{1} \times 0 + b_{1} \times 1 = b_{1}

c_{3} = a_{2} \times 1 + b_{2} \times 0 = a_{2}

c_{4} = a_{2} \times 0 + b_{2} \times 1 = b_{2}

⋮

c_{2 n - 2} = a_{n - 1} \times 0 + b_{n - 1} \times 1 = b_{n - 1}

c_{2 n - 1} = a_{n} \times 1 + b_{n} \times 0 = a_{n}

c_{2 n} = a_{n} \times 0 + b_{n} \times 1 = b_{n}

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