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Gujarati JEE Mathematics : દ્વિપદી પ્રમેય

Multiple Choice Questions

21. જો $bold sum from bold r bold equals bold 0 to bold n of bold thin space bold left parenthesis bold minus bold 1 bold right parenthesis to the power of bold r bold space bold left parenthesis bold a bold minus bold r bold right parenthesis bold space open parentheses table row bold n row bold r end table close parentheses bold space bold equals bold space bold. bold. bold. bold. bold. bold space bold.$

a + 1
na
a
0

Tips: -

$bold sum from bold r bold equals bold 0 to bold n of bold space bold left parenthesis bold minus bold 1 bold right parenthesis to the power of bold r bold space bold left parenthesis bold a bold minus bold r bold right parenthesis bold space open parentheses table row bold n row bold r end table close parentheses bold space bold equals bold left parenthesis bold a bold minus bold 0 bold right parenthesis bold space open parentheses table row bold n row bold 0 end table close parentheses bold space bold minus bold space bold left parenthesis bold a bold space bold minus bold space bold 1 bold right parenthesis bold space open parentheses table row bold n row bold 1 end table close parentheses bold space bold plus bold space bold left parenthesis bold a bold space bold minus bold space bold 2 bold right parenthesis bold space open parentheses table row bold n row bold 2 end table close parentheses bold space bold plus bold space bold. bold. bold. bold space bold plus bold space bold left parenthesis bold minus bold 1 bold right parenthesis to the power of bold n bold space bold left parenthesis bold a bold minus bold n bold right parenthesis bold space open parentheses table row bold n row bold n end table close parentheses 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 equals bold space bold a bold space open square brackets open parentheses table row bold n row bold 0 end table close parentheses bold minus open parentheses table row bold n row bold 1 end table close parentheses bold plus open parentheses table row bold n row bold 2 end table close parentheses bold minus bold space bold. bold. bold. bold space bold plus bold space bold left parenthesis bold minus bold 1 bold right parenthesis to the power of bold n bold space open parentheses table row bold n row bold n end table close parentheses close square brackets bold space bold plus bold space open square brackets open parentheses table row bold n row bold 1 end table close parentheses bold minus bold 2 bold space open parentheses table row bold n row bold 2 end table close parentheses bold space bold plus bold space bold 3 bold space open parentheses table row bold n row bold 3 end table close parentheses bold space bold minus bold space bold. bold. bold. bold space bold plus bold space bold left parenthesis bold minus bold 1 bold right parenthesis to the power of bold n subscript bold n bold space open parentheses table row bold n row bold n end table close parentheses close square brackets bold space bold space bold. bold. bold. bold space bold left parenthesis bold i bold right parenthesis$

હવે, $bold left parenthesis bold 1 bold space bold minus bold space bold x bold right parenthesis to the power of bold n bold space bold equals bold space open parentheses table row bold n row bold 0 end table close parentheses bold space bold minus bold space open parentheses table row bold n row bold 1 end table close parentheses bold space bold x bold space bold plus bold space open parentheses table row bold n row bold 2 end table close parentheses bold space bold x to the power of bold 2 bold space bold minus bold space bold. bold. bold. bold space bold plus bold space bold left parenthesis bold minus bold 1 bold right parenthesis to the power of bold n bold space open parentheses table row bold n row bold n end table close parentheses bold space bold x to the power of bold n bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold. bold. bold. bold space bold left parenthesis bold ii bold right parenthesis bold space$

x = 1 લેતાં, $bold 0 bold space bold equals bold space open parentheses table row bold n row bold 0 end table close parentheses bold space bold minus bold space open parentheses table row bold n row bold 1 end table close parentheses bold space bold plus bold space open parentheses table row bold n row bold 2 end table close parentheses bold space bold minus bold space bold. bold. bold. bold space bold plus bold space bold left parenthesis bold minus bold 1 bold right parenthesis to the power of bold n bold space open parentheses table row bold n row bold n end table close parentheses bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold. bold. bold. bold space bold left parenthesis bold iii bold right parenthesis$
સમીકરણ (ii) નું બંને બાજુ x ને સાપેક્ષ વિકલન કરતાં

$bold minus bold n bold space bold left parenthesis bold 1 bold space bold minus bold space bold x bold right parenthesis to the power of bold n bold minus bold 1 end exponent bold space bold equals bold space bold minus open parentheses table row bold n row bold 1 end table close parentheses bold space bold plus bold space bold 2 bold space open parentheses table row bold n row bold 2 end table close parentheses bold space bold x bold space bold minus bold space bold 3 bold x to the power of bold 2 bold space open parentheses table row bold n row bold 3 end table close parentheses bold space bold plus bold space bold. bold. bold. bold space bold plus bold space bold left parenthesis bold minus bold 1 bold right parenthesis to the power of bold n bold space bold nx to the power of bold n bold minus bold 1 end exponent bold space open parentheses table row bold n row bold n end table close parentheses$

x = 1 લેતાં, $bold 0 bold space bold equals bold space bold minus bold space open parentheses table row bold n row bold 1 end table close parentheses bold space bold plus bold space bold 2 bold space open parentheses table row bold n row bold 2 end table close parentheses bold space bold minus bold space bold 3 bold space open parentheses table row bold n row bold 3 end table close parentheses bold space bold plus bold space bold. bold. bold. bold space bold plus bold space bold left parenthesis bold minus bold 1 bold right parenthesis to the power of bold n bold space bold n bold space open parentheses table row bold n row bold n end table close parentheses$

$bold therefore bold space bold 0 bold space bold equals bold space open parentheses table row bold n row bold 1 end table close parentheses bold space bold minus bold space bold 2 bold space open parentheses table row bold n row bold 2 end table close parentheses bold space bold plus bold space bold 3 bold space open parentheses table row bold n row bold 3 end table close parentheses bold space bold minus bold space bold. bold. bold. bold space bold minus bold space bold left parenthesis bold minus bold 1 bold right parenthesis to the power of bold n bold space bold b bold space open parentheses table row bold n row bold n end table close parentheses$ ... (iv)

પરિણામ (i), (iii), (iv) પરથી

$bold sum from bold r bold equals bold 0 to bold n of bold space bold left parenthesis bold minus bold 1 bold right parenthesis to the power of bold r bold space bold left parenthesis bold a bold space bold minus bold r bold right parenthesis bold space open parentheses table row bold n row bold r end table close parentheses bold space bold equals bold space bold a bold space bold left parenthesis bold 0 bold right parenthesis bold space bold plus bold space bold 0 bold space bold equals bold space bold 0 bold space$

22. જો (1 + x)¹⁰⁰ = c₀ + c₁x + c_zx² + ... + c₁₀₀x¹⁰⁰ તો $bold 2 to the power of bold 1 bold space bold c subscript bold 0 over bold 1 bold space bold plus bold space bold 2 to the power of bold 2 bold space bold c subscript bold 1 over bold 2 bold space bold plus bold space bold 2 to the power of bold 2 bold space bold c subscript bold 2 over bold 3 bold space bold plus bold space bold. bold. bold. bold space bold plus bold space bold 2 to the power of bold 101 bold space bold c subscript bold 100 over bold 101 bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold space$

$bold 3 to the power of bold 101 over bold 101$
$fraction numerator bold 3 to the power of bold 101 bold plus bold 1 over denominator bold 101 end fraction$
$fraction numerator bold 3 to the power of bold 101 bold minus bold 1 over denominator bold 101 end fraction$
$fraction numerator bold 2 to the power of bold 101 bold minus bold 1 over denominator bold 101 end fraction$

23.

એક પરીક્ષક કોઈપણ પ્રશ્નમાં 2 થી ઓછા ગુણ આપ્યા સિવાય, 8 પ્રશ્નમાં 25 ગુણ કેટલી રીતે આપી શકે ?

$open parentheses table row bold 16 row bold 8 end table close parentheses$
$open parentheses table row bold 17 row bold 9 end table close parentheses$
$open parentheses table row bold 16 row bold 7 end table close parentheses$
₁₆P₇

24. 12 સમાન પેનો 5 વિદ્યાર્થીઓને કુલ કેટલી વહેંચી શકાય ?

$open parentheses table row bold 16 row bold 5 end table close parentheses$
$open parentheses table row bold 11 row bold 4 end table close parentheses$
$open parentheses table row bold 16 row bold 4 end table close parentheses$
₁₆P₄

25.

અસંખ્ય લાલ, સફેદ, કાળા અને લીલા દડાઓમાંથી દરેક રંગના દડા આવે તે રીતે 11 દડા પસંદ કરવાના પ્રકાર કેટલા થાય ?

165
330
120
અસંખ્ય

26. જો $open parentheses table row cell bold n bold plus bold 1 end cell row bold 1 end table close parentheses bold space open parentheses table row cell bold n bold plus bold 1 end cell row bold 2 end table close parentheses bold space open parentheses table row cell bold n bold plus bold 1 end cell row bold n end table close parentheses bold space bold equals bold space bold. bold. bold. bold. bold space bold. bold space$

$fraction numerator bold left parenthesis bold n bold plus bold 1 bold right parenthesis to the power of bold n over denominator bold n bold space bold factorial end fraction open parentheses table row bold n row bold 1 end table close parentheses bold space open parentheses table row bold n row bold 2 end table close parentheses bold space bold. bold. bold. bold space open parentheses table row bold n row bold n end table close parentheses$
$fraction numerator bold left parenthesis bold n bold minus bold 1 bold right parenthesis to the power of bold n over denominator bold n bold space bold factorial end fraction open parentheses table row bold n row bold 1 end table close parentheses bold space open parentheses table row bold n row bold 2 end table close parentheses bold space bold. bold. bold. bold space open parentheses table row bold n row bold n end table close parentheses$
$bold n to the power of bold n bold space open parentheses table row bold n row bold 1 end table close parentheses bold space open parentheses table row bold n row bold 2 end table close parentheses bold space bold. bold. bold. bold space open parentheses table row bold n row bold n end table close parentheses$
$bold left parenthesis bold n bold plus bold 1 bold right parenthesis to the power of bold n bold space open parentheses table row bold n row bold 1 end table close parentheses bold space open parentheses table row bold n row bold 2 end table close parentheses bold space bold. bold. bold. bold space open parentheses table row bold n row bold n end table close parentheses$

27.

એક રૂપિયાના 20 સિક્કા, બે રૂપિયાના 15 સિક્કા અને પાંચ રૂપિયાના 10 સિક્કામાંથી 4 સિક્કા કુલ કેટલી રીતે પસંદ કરી શકાય ?

28. જો (1 + x)ⁿ = c₀ + c₁x + c₂x² + ... + c_nxⁿ તો c₀ - $bold 1 over bold 2 bold space bold c subscript bold 1 bold plus bold space bold 1 over bold 3 bold space bold c subscript bold 2 bold space bold minus bold space bold. bold. bold. bold space bold plus bold space bold left parenthesis bold minus bold 1 bold right parenthesis to the power of bold n bold space fraction numerator bold 1 over denominator bold n bold plus bold 1 end fraction bold space bold c subscript bold n bold space bold equals bold space bold. bold. bold. bold. bold. bold space$

$fraction numerator bold 1 over denominator bold n bold plus bold 1 end fraction$
$fraction numerator bold 2 to the power of bold n bold plus bold 1 end exponent bold minus bold 1 over denominator bold n bold plus bold 1 end fraction$
$fraction numerator bold 2 to the power of bold n bold plus bold 1 end exponent bold plus bold 1 over denominator bold n bold plus bold 1 end fraction$
$fraction numerator bold 2 over denominator bold n bold plus bold 1 end fraction$

29.

ગણિત, ભૌતિક અને રસાયણ વિષયનાં પુસ્તકોના સમૂહમાં દરેક વિષયનાં ઓછામાં ઓછા 10 પુસ્તકો (ધોરણ-12ના) છે. આ સમુહમાંથી 5 પુસ્તકો કેટલી રીતે પસંદ કરી શકાય ?

30.

એક વિદ્યાર્થી ત્રણ વિષયની પરીક્ષા આપે છે. દરેક વિષયના પેપર મહત્તમ n ગુણના છે. તે 2n ગુણ કેટલી રીતે મેળવી શકે ?

$fraction numerator bold n bold left parenthesis bold n bold plus bold 1 bold right parenthesis over denominator bold 2 end fraction$
$open parentheses table row cell bold 3 bold n end cell row cell bold 2 bold n end cell end table close parentheses$
n(n+1)(n+2)
$fraction numerator bold left parenthesis bold n bold plus bold 1 bold right parenthesis bold left parenthesis bold n bold plus bold 2 bold right parenthesis over denominator bold 2 end fraction$

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Gujarati JEE Mathematics : દ્વિપદી પ્રમેય

Multiple Choice Questions

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