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Class 10 Class 12

ગણિતિય અનુમાનો સિદ્વાંત

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

વિધાન p(n) : n²≥ 3n ........ સત્ય છે.

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

$bold p bold left parenthesis bold n bold right parenthesis bold space bold colon bold space bold 2 to the power of bold 2 to the power of bold 2 end exponent$ 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 = .......... .

Tips: -

$bold p bold left parenthesis bold 1 bold right parenthesis bold space bold colon bold space bold 1 over bold 3 bold space bold less than bold space bold 1 bold space bold less than bold space bold 1 over bold 2$ સત્ય નથી.

$bold p bold left parenthesis bold 2 bold right parenthesis bold thin space bold colon bold space bold 4 over bold 9 bold space bold less than bold space bold 2 bold space bold less than bold space bold 1$ સત્ય નથી.

$bold p bold left parenthesis bold 3 bold right parenthesis bold space bold colon bold space bold 1 bold space bold less than bold space bold 6 bold space bold less than bold space bold 27 over bold 8$ સત્ય નથી.

$bold p bold left parenthesis bold 4 bold right parenthesis bold space bold colon bold space bold 256 over bold 81 bold space bold less than bold space bold 24 bold space bold less than bold space bold 256 over bold 16$ સત્ય નથી.

$bold p bold left parenthesis bold 5 bold right parenthesis bold space bold colon bold space bold 5 to the power of bold 5 over bold 3 to the power of bold 5 bold space bold less than bold space bold 120 bold space bold less than bold space bold 5 to the power of bold 5 over bold 2 to the power of bold 5$ એટલે કે $bold 3125 over bold 243 bold less than bold 120 bold space bold less than bold space bold 3125 over bold 32$ સત્ય નથી.

$bold p bold left parenthesis bold 6 bold right parenthesis bold space bold colon bold space bold 6 to the power of bold 3 over bold 3 to the power of bold 6 bold space bold less than bold space bold 6 bold factorial bold space bold less than bold space bold 6 to the power of bold 6 over bold 2 to the power of bold 6$ એટલે કે $bold 6 to the power of bold 5 over bold 3 to the power of bold 6 bold space bold less than bold 5 bold factorial bold space bold less than bold space bold 6 to the power of bold 5 over bold 3 to the power of bold 6 bold space bold less than bold space bold 5 bold factorial$ એટલે કે $bold 2 to the power of bold 5 over bold 3 bold space bold less than bold space bold 120 bold space bold less than bold space bold 3 to the power of bold 5 over bold 2$

એટલે કે $bold 32 over bold 2 bold space bold less than bold space bold 120 bold space bold less than bold space bold 243 over bold 2$ સત્ય છે.

આ જ રીતે, સાબિત કરી શકાય કે p(n) દરેક n > 6 માટે સત્ય છે.

અહીં ગણિતીય અનુમાનના સિદ્વાંત પરથી $bold n bold space bold factorial bold thin space bold less than bold space bold n to the power of bold n over bold 2 to the power of bold n$ અને $bold n to the power of bold n over bold 3 to the power of bold n bold space bold less than bold space bold n bold space bold factorial bold comma bold space bold n bold space bold greater or equal than bold space bold 6$ સાબિત કરી શકાય.

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

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ગણિતિય અનુમાનો સિદ્વાંત

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