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  # सिद्ध कीजिए कि - <br> (i) 1^ 2 * C_ 1 + 2 ^ 2 * C _ 2 + 3 ^ 2 * C_ 3 + ... + n^ 2 * C _ n = n ( n+ 1 ) 2 ^ (n - 2 ) <br> (ii) a C _ 0 + ( a + b ) C _ 1 + ( a + 2b ) C _ 2 + ... + ( a + nb ) C _ n = ( 2a + nb ) 2 ^ (n- 1 ) <br> (iii) C _ 3 + 2 C _ 4 + 3C_ 5 + ... + (n - 2 ) C _ n = (n - 4 ) 2^ ( n - 1 ) + n + 2 , जहाँ n gt 3 <br> (iv) C _ 0 - C _ 1 + C _ 2 - C_ 3 + ... + ( -1 ) ^nC _ n = 0
  
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  Solution :
  (i)  1 ^ 2 * C _ 1 + 2 ^ 2 * C _ 2 + 3 ^ 2 * C _ 3 + … + n^ 2 * C_ n  <br>  = sum _ ( r = 1 ) ^ n r ^ 2 C_ r = sum _ ( r= 1 ) ^ n r ^ 2 ""^ n C _ r = sum _ ( r = 1 ) ^ n [r ( r - 1 ) + r] ""^ nC _ r  <br>  = sum _ ( r = 1 ) ^ n r ( r- 1 ) ""^n C _ r + sum _ ( r = 1 ) ^ n r * ""^ n C _ r  <br>  = sum _ ( r = 2 ) ^n r * (r -1 ) (n ) / ( r ) ( ( n-1 ) )/( (r - 1 )) ""^ (n - 2 ) C _ ( r - 2 ) + sum _ (r = 1 ) ^n r * ( n ) /( r ) ""^ (n- 1 ) C _ ( r - 1 )  <br>  = n ( n - 1 ) ( sum _ ( r = 2 )^n ""^ (n - 2 ) C_ ( r- 2 ) ) + n ( sum - ( r=1)^ n ""^ (n- 1 ) C_ ( r - 1 ))  <br>  = n (n - 1) [""^ ( n -2 )C_ 0 +""^ ( n -2 ) C_ 1 + ... + ""^ (n - 2 ) C_ ( n - 2 ) ] + n[ ""^ (n - 1 ) C_ 0 + ""^ (n - 1 )C_ 1 + ... + ""^ (n - 1 ) C_ (n - 1 ) ]  <br> (ii)  a C _ 0 + ( a + b ) C_ 1 + (a + 2b ) C_ 2 + ... + ( a + n b ) C _ n  <br>  = sum _ ( r = 0 ) ^n ( a + r b ) ""^ nC _ r = sum _ ( r = 0 ) ^ n a * ""^ n C _ r + sum _ ( r = 0 ) ^n rb""^ n C_ r = a ( sum _ ( r= 0 ) ^n ""^n C _ r ) + b ( sum _ ( r = 0 ) ^n r * ""^ nC _ r )  <br>  = a ( sum _ ( r = 0 ) ^ n ""^ n C_ r ) + b ( sum_ ( r = 1 )^ n r * ( n)/( r ) ""^ (n - 1 ) C_ ( r - 1 ) ) = a ( sum _ ( r=0 )^n ""^n C _ r ) + bn ( sum _ ( r = 1 ) ^ n ""^ (n - 1 ) C _ ( r - 1 ) )  <br>  = a2^n + bn2^(n- 1 ) " " [ because sum _ ( r = 0 ) ^ n ""^ n C_ r = 2^n , sum _ ( r = 1 ) ^n ""^ ( n - 1 ) C _ ( r - 1 ) = 2 ^ (n - 1 ) ]  <br>  = ( 2a + bn ) 2 ^ (n - 1 )  <br> (iii)  C _ 3 + 2C _ 4 + 3C _ 5 + ... + (n - 2 ) C_ n = sum _ ( r = 3 ) ^n ( r - 2 ) C_ r  <br>  = sum_ ( r = 3 ) ^n ( r- 2 ) ""^ n C _ r = sum _ ( r = 3 ) ^ n r * ""^ n C_ r = sum _ ( r = 3 ) ^n r * ( n ) /( r ) ""^ (n- 1 ) C _ ( r - 1 ) -2 sum_ ( r = 3 ) ^ n ""^ n C _ r  <br>  = n ( sum _ ( r = 3 ) ^ n ""^ (n - 1 ) C_ ( r -1 )) - 2 ( sum _ ( r = 3 )^ n ""^ n C _ r )  <br>  = n ( ""^ ( n - 1 ) C_ 2 + ""^ (n - 1 ) C _ 3 + ... + ""^ (n - 1 ) C _ ( n - 1 )) - 2 ( ""^ nC _ 3 + ""^n C _ 4 + ... + ""^ n C _ n )  <br>  = n[ ( ""^ ( n - 1 ) C_ 0 + ""^ (n - 1 ) C _ 1 + ""^ (n - 1 ) C _ 2 ... + ""^ ( n - 1 ) C _ (n - 1 ) ) - ( ""^ ( n - 1 ) C _ 0 + ""^ (n- 1 ) C_ 1 ) ]  <br>  -2[ ( 2^n C _ 0 + ""^n C_ 1 + ""^ n C _ 2 + ""^ n C_ 3 + ... + ""^n C _n ) - ( ""^n C _ 0 + ""^n C _ 1 + ""^n C _ 2 ) ]  <br>  = n [ 2 ^ ( n - 1 ) - { 1 + ( n - 1 ) }] - 2[ 2^n - ( 1+ n + (n ( n- 1 )) / ( 2 ) ) ]  <br>  = n[ 2 ^ (n - 1 ) - n ] - 2[ 2^n - ((n^ 2 + n + 2 ) /( 2)) ]  <br>  = n*2 ^ (n - 1 ) - n^ 2 - 2 * 2 ^ n + n^ 2 + n + 2 = n* 2 ^ (n - 1 ) - 2^(n + 1 ) + n + 2  <br>  = (n - 4 ) 2 ^ (n - 1 ) + n + 2  <br> (vi)  ( 1 + x ) ^n = C_ 0 + C _ 1 x + C _ 2 x ^ 2 + C_ 3 x ^ 3 + ...+C_ n x ^n  <br>  x = - 1  रखने पर ,  C _ 0 - C _ 1 + C _ 2 - C _ 3 + … + (-1 ) ^n C_ n = 0
  
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explanation: The document is a math problem set in Hindi, specifically from a chapter on the Binomial Theorem. It contains multiple questions that require proving certain binomial identities. Each question is followed by a detailed step-by-step solution.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: Home
  
  >
  
  Hindi
  
  >
  
  Class 11
  
  >
  
  Maths
  
  >
  
  Chapter
  
  >
  
  द्विपद प्रमेय
  
  >
  
  सिद्ध कीजिए कि - <br> (i)...
  
  # सिद्ध कीजिए कि - <br> (i) 1^ 2 * C_ 1 + 2 ^ 2 * C _ 2 + 3 ^ 2 * C_ 3 + ... + n^ 2 * C _ n = n ( n+ 1 ) 2 ^ (n - 2 ) <br> (ii) a C _ 0 + ( a + b ) C _ 1 + ( a + 2b ) C _ 2 + ... + ( a + nb ) C _ n = ( 2a + nb ) 2 ^ (n- 1 ) <br> (iii) C _ 3 + 2 C _ 4 + 3C_ 5 + ... + (n - 2 ) C _ n = (n - 4 ) 2^ ( n - 1 ) + n + 2 , जहाँ n gt 3 <br> (iv) C _ 0 - C _ 1 + C _ 2 - C_ 3 + ... + ( -1 ) ^nC _ n = 0
  
  Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
  Apne doubts clear karein ab Whatsapp par bhi. Try it now.
  
  Watch 1000+ concepts & tricky questions explained!
  
  5.6 K+
  
  200+
  
  Text Solution
  
  Solution :
  (i)  1 ^ 2 * C _ 1 + 2 ^ 2 * C _ 2 + 3 ^ 2 * C _ 3 + … + n^ 2 * C_ n  <br>  = sum _ ( r = 1 ) ^ n r ^ 2 C_ r = sum _ ( r= 1 ) ^ n r ^ 2 ""^ n C _ r = sum _ ( r = 1 ) ^ n [r ( r - 1 ) + r] ""^ nC _ r  <br>  = sum _ ( r = 1 ) ^ n r ( r- 1 ) ""^n C _ r + sum _ ( r = 1 ) ^ n r * ""^ n C _ r  <br>  = sum _ ( r = 2 ) ^n r * (r -1 ) (n ) / ( r ) ( ( n-1 ) )/( (r - 1 )) ""^ (n - 2 ) C _ ( r - 2 ) + sum _ (r = 1 ) ^n r * ( n ) /( r ) ""^ (n- 1 ) C _ ( r - 1 )  <br>  = n ( n - 1 ) ( sum _ ( r = 2 )^n ""^ (n - 2 ) C_ ( r- 2 ) ) + n ( sum - ( r=1)^ n ""^ (n- 1 ) C_ ( r - 1 ))  <br>  = n (n - 1) [""^ ( n -2 )C_ 0 +""^ ( n -2 ) C_ 1 + ... + ""^ (n - 2 ) C_ ( n - 2 ) ] + n[ ""^ (n - 1 ) C_ 0 + ""^ (n - 1 )C_ 1 + ... + ""^ (n - 1 ) C_ (n - 1 ) ]  <br> (ii)  a C _ 0 + ( a + b ) C_ 1 + (a + 2b ) C_ 2 + ... + ( a + n b ) C _ n  <br>  = sum _ ( r = 0 ) ^n ( a + r b ) ""^ nC _ r = sum _ ( r = 0 ) ^ n a * ""^ n C _ r + sum _ ( r = 0 ) ^n rb""^ n C_ r = a ( sum _ ( r= 0 ) ^n ""^n C _ r ) + b ( sum _ ( r = 0 ) ^n r * ""^ nC _ r )  <br>  = a ( sum _ ( r = 0 ) ^ n ""^ n C_ r ) + b ( sum_ ( r = 1 )^ n r * ( n)/( r ) ""^ (n - 1 ) C_ ( r - 1 ) ) = a ( sum _ ( r=0 )^n ""^n C _ r ) + bn ( sum _ ( r = 1 ) ^ n ""^ (n - 1 ) C _ ( r - 1 ) )  <br>  = a2^n + bn2^(n- 1 ) " " [ because sum _ ( r = 0 ) ^ n ""^ n C_ r = 2^n , sum _ ( r = 1 ) ^n ""^ ( n - 1 ) C _ ( r - 1 ) = 2 ^ (n - 1 ) ]  <br>  = ( 2a + bn ) 2 ^ (n - 1 )  <br> (iii)  C _ 3 + 2C _ 4 + 3C _ 5 + ... + (n - 2 ) C_ n = sum _ ( r = 3 ) ^n ( r - 2 ) C_ r  <br>  = sum_ ( r = 3 ) ^n ( r- 2 ) ""^ n C _ r = sum _ ( r = 3 ) ^ n r * ""^ n C_ r = sum _ ( r = 3 ) ^n r * ( n ) /( r ) ""^ (n- 1 ) C _ ( r - 1 ) -2 sum_ ( r = 3 ) ^ n ""^ n C _ r  <br>  = n ( sum _ ( r = 3 ) ^ n ""^ (n - 1 ) C_ ( r -1 )) - 2 ( sum _ ( r = 3 )^ n ""^ n C _ r )  <br>  = n ( ""^ ( n - 1 ) C_ 2 + ""^ (n - 1 ) C _ 3 + ... + ""^ (n - 1 ) C _ ( n - 1 )) - 2 ( ""^ nC _ 3 + ""^n C _ 4 + ... + ""^ n C _ n )  <br>  = n[ ( ""^ ( n - 1 ) C_ 0 + ""^ (n - 1 ) C _ 1 + ""^ (n - 1 ) C _ 2 ... + ""^ ( n - 1 ) C _ (n - 1 ) ) - ( ""^ ( n - 1 ) C _ 0 + ""^ (n- 1 ) C_ 1 ) ]  <br>  -2[ ( 2^n C _ 0 + ""^n C_ 1 + ""^ n C _ 2 + ""^ n C_ 3 + ... + ""^n C _n ) - ( ""^n C _ 0 + ""^n C _ 1 + ""^n C _ 2 ) ]  <br>  = n [ 2 ^ ( n - 1 ) - { 1 + ( n - 1 ) }] - 2[ 2^n - ( 1+ n + (n ( n- 1 )) / ( 2 ) ) ]  <br>  = n[ 2 ^ (n - 1 ) - n ] - 2[ 2^n - ((n^ 2 + n + 2 ) /( 2)) ]  <br>  = n*2 ^ (n - 1 ) - n^ 2 - 2 * 2 ^ n + n^ 2 + n + 2 = n* 2 ^ (n - 1 ) - 2^(n + 1 ) + n + 2  <br>  = (n - 4 ) 2 ^ (n - 1 ) + n + 2  <br> (vi)  ( 1 + x ) ^n = C_ 0 + C _ 1 x + C _ 2 x ^ 2 + C_ 3 x ^ 3 + ...+C_ n x ^n  <br>  x = - 1  रखने पर ,  C _ 0 - C _ 1 + C _ 2 - C _ 3 + … + (-1 ) ^n C_ n = 0
  
  121321986
  
  200+
  
  5.6 K+
  
  121321985
  
  200+
  
  4.2 K+
  
  76129482
  
  1.7 K+
  
  34.9 K+
  
  4:31
  121321987
  
  200+
  
  4.8 K+
  
  88356926
  
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  28.9 K+
  
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  1.7 K+
  
  34.4 K+
  
  4:10
  104444775
  
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  23.0 K+
  
  4:37
  76129422
  
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  88356927
  
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  13.8 K+
  
  131221363
  
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  127117107
  
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distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a math problem set in Hindi, specifically from a chapter on the Binomial Theorem. It contains multiple questions that require proving certain binomial identities. Each question is followed by a detailed step-by-step solution.
  
  Label: multiple-question-and-answer