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CBSE

Gujarati JEE Mathematics : ગણિતિય અનુમાનો સિદ્વાંત

Multiple Choice Questions

1. જો $bold sum from bold r bold equals bold 1 to bold k of bold r bold space bold equals bold space bold 1 over bold 2 bold space bold left parenthesis bold n to the power of bold 2 bold space bold plus bold space bold 11 bold n bold space bold plus bold space bold 30 bold right parenthesis$ તો k =- ....... .

n + 6
n + 7
n + 5
n + 4

2. જો 1³ + 2³ + 3³ + ... + 50³ = m², તો m = ........... .

1275
1225
2450
1375

3. 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 = .......... .

bold n bold factorial bold space bold less than bold space open square brackets fraction numerator bold n bold plus bold 1 over denominator bold 2 end fraction close square brackets to the power of bold n

નું પાલન થાય તેવો નાનામાં નાનો ધનપૂર્ણાંક ........ છે. જ્યાં [] મહત્તમ પૂર્ણાંક વિધેય છે.

5. વિધાન p(n) : n²≥ 3n ........ સત્ય છે.

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

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

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

Tips: -

n = 2 લેતાં $bold 2 to the power of bold 2 to the power of bold 2 end exponent bold space bold equals bold space bold 16$ નો એકમનો અંક 6 છે.

n = 3 લેતાં $bold 2 to the power of bold 2 to the power of bold 3 end exponent bold space bold equals bold space bold 2 to the power of bold 8 bold space bold space bold equals bold space bold 256$ નો એકમનો અંક 6 છે.

∴ અનુમાન કરી શકાય કે દરેક n > 1 માટે $bold 2 to the power of bold 2 to the power of bold n end exponent$ નો એકમનો અંક 6 છે.
ધારો કે p(x) : $bold 2 to the power of bold 2 to the power of bold k end exponent$ નો એકમનો અંક 6 છે. k > 1

n = k + 1 લેતાં, $bold 2 to the power of bold 2 to the power of bold k bold plus bold 1 end exponent end exponent bold space bold equals bold space bold 2 to the power of bold 2 to the power of bold k end exponent bold space bold space bold 2 to the power of bold 2$

$bold equals bold space open parentheses bold 2 to the power of bold 2 to the power of bold k end exponent close parentheses to the power of bold 2 bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space open parentheses bold 2 to the power of bold 2 to the power of bold k end exponent bold space bold equals bold space bold 10 bold m bold space bold plus bold space bold 6 bold comma bold space bold m bold element of bold space bold N close parentheses bold equals bold space bold left parenthesis bold 10 bold space bold m bold space bold plus bold space bold 6 bold right parenthesis to the power of bold 2 bold equals bold space bold 100 bold space bold m to the power of bold 2 bold space bold plus bold space bold 120 bold m bold space bold plus bold space bold 36 bold space bold equals bold space bold 100 bold space bold m to the power of bold 2 bold space bold plus bold space bold 120 bold m bold space bold plus bold space bold 30 bold space bold plus bold space bold 6 bold space$

$bold equals bold space bold 10 bold space bold left parenthesis bold 10 bold space bold italic m to the power of bold 2 bold space bold plus bold space bold 12 bold italic m bold space bold plus bold space bold 3 bold right parenthesis bold space bold plus bold space bold 6 bold space$ જેનો એકમનો અંક 6 છે.
∴ p(k + 1) સત્ય છે.

આમ p(k) સત્ય છે. ⇒ p(k+1) સત્ય છે. k > 1

∴ p(n) દરેક n > 1 માટે સત્ય છે.