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If $a_{1}$ = 1 and $a_{n + 1}$ – 3 $a_{n}$ + 2 = 4n for every positive integer n, then $a_{100}$ equals

$3^{99}$ – 200

$3^{99}$ + 200

$3^{100}$ – 200

$3^{100}$ + 200

View Solution

$a_{1}$ = 1

$a_{n + 1}$ = 4n + 3 $a_{n}$ – 2

$a_{2}$ = 4 + 3(1) – 2 = 5 = $3^{2}$ – 4

$a_{3}$ = 4(2) + 3(5) – 2 = 21 = $3^{3}$ – 6

$a_{4}$ = 4(3) + 3(21) – 2 = 73 = $3^{4}$ – 8

∴ $a_{n}$ = 3n – 2(n)

∴ $a_{100}$ = $3^{100}$ – 200

Hence, option 3.

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Ramanjaneyulu s

Thank you

Sudhakar

Thans For Giving Solution

Pavan Jaiswal

This Solution is very simple thanks for giving video solution.its very eay for understood.