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ગણિતિય અનુમાનો સિદ્વાંત

વિધાન p(n) : 3²ⁿ⁺¹+2^n-1એ n ∈ N .......... ના ગુણકમાં છે.

$bold p bold left parenthesis bold n bold right parenthesis bold space bold colon bold space bold 2 to the power of bold 2 to the power of bold 2 end exponent$ નો એકમનો અંક ....... છે. n > 1

વિધાન $bold p bold left parenthesis bold n bold right parenthesis bold space bold colon bold space bold n to the power of bold 3 over bold 3 to the power of bold n bold space bold less than bold space bold n bold factorial bold space bold less than bold space bold n to the power of bold n over bold 2 to the power of bold n$ દરેક n ≥ k, n ∈ N માટે સત્ય છે, તો k = .......... .

વિધાન p(n) : n²≥ 3n ........ સત્ય છે.

n ∈ N
n ∈ N, n > 1
n ∈ N, n ≥ 3
દરેક અયુગ્મ પ્રાકૃતિક સંખ્યા

જો $bold a subscript bold n bold space bold equals bold space square root of bold 7 bold plus square root of bold 7 bold plus square root of bold 7 bold plus bold. bold. bold. end root end root bold space bold n$ વખત તો ગણિતીય અનુમાનના સિદ્વાંત પરથી કયું સત્ય છે?

a_n > 7, n ≥ 1
a_n > 3, n ≥ 1
a_n < 4, n ≥ 1
a_n < 3, n ≥ 1

a_n < 4, n ≥ 1

Tips: -

$bold p bold left parenthesis bold n bold right parenthesis bold space bold colon bold space bold a subscript bold n bold space bold equals bold space square root of bold 7 bold plus square root of bold 7 square root of bold 7 bold plus bold. bold. bold. end root end root end root bold space bold n$ વખત

n = 1લેતાં, a₁ = $square root of bold 7$ આથી a₁ < 7 હોવાથી a_n > 7 સત્ય નથી.

a₁ > 3 સત્ય નથી. a₁ = $square root of bold 7 bold space bold greater than bold space bold 4$ અને a₁ = $square root of bold 7$ < 3 સત્ય છે.

$bold a subscript bold 2 bold space bold equals bold space square root of bold 7 bold plus square root of bold 7 end root$
અહીં $bold 2 bold space bold less than bold space square root of bold 7 bold space bold less than bold space bold 3 bold space bold rightwards double arrow bold space bold 9 bold space bold less than bold space bold 7 bold space bold plus bold space square root of bold 7 bold space bold less than bold space bold 10 bold space bold less than bold space bold 10 bold space bold rightwards double arrow bold space bold 3 bold space bold less than bold space square root of bold 7 square root of bold 7 end root bold space bold less than bold space bold 4 bold space bold rightwards double arrow bold space bold 3 bold space bold less than bold space bold a subscript bold 2 bold space bold less than bold space bold 4$

∴ a₂ < 3 સત્ય નથી. પરંતુ a₂ < 4 સત્ય છે.

આમ, a₁ < 4, a₂ < 4 સત્ય છે.

ધારો કે a_k < 4

હવે, a_k+1 = $square root of bold 7 bold plus square root of bold 7 bold plus square root of bold 7 bold plus bold. bold. bold. end root end root end root bold space bold k bold space bold plus bold space bold 1$ વખત

$bold equals bold space square root of bold 7 bold plus bold a subscript bold k end root$

$bold therefore bold space bold a to the power of bold 2 subscript bold k bold plus bold 1 end subscript bold space bold equals bold space bold 7 bold space bold plus bold space bold a subscript bold k bold space bold less than bold space bold 7 bold space bold plus bold space bold 4 bold.$ આથી $bold a subscript bold k bold plus bold 1 end subscript bold space bold less than bold space square root of bold 11 bold space bold less than bold space bold 4$ આથી $bold a subscript bold k bold plus bold 1 end subscript bold space bold less than bold space bold 4$

આમ, a_n <4, n = k+1 માટે સત્ય છે.

$bold therefore bold space bold a subscript bold n bold space bold less than bold space bold 4 bold comma bold space$ n ∈ N

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