📖 generic · CBSE Class 11 English medium · MATHEMATICS · Page 177question

Prove that ∑ · Part 5

Chapter 2: 4 P 2 = 4 C 2 ××××× 2! or ( · MATHEMATICS

Solution We have T r + = C C    MATHEMATICS ( ) ) C ( ) The term will be independent of x if the index of x is zero, i.e., – r = . Thus, r = Hence th term is independent of x and is given by (– ) C ) Example If the coefficients of a r – , a r and a r + in the expansion of ( + a ) n are in arithmetic progression, prove that n – n ( r + ) + r – = . Solution The ( r + ) th term in the expansion is n C r a r . Thus it can be seen that a r occurs in the ( r + ) th term, and its coefficient is n C r .

Hence the coefficients of a r – , a r and a r + are n C r – , n C r and n C r + , respectively. Since these coefficients are in arithmetic progression, so we have, n C r – + n C r + = . n C r . This gives )!

( )! ( – ) ( ) ) ( )     −+   )! ( )![ ( – )] r n × i.e. ) () , or ) )( ) )( ) ( ) r r r r r n or r ( r + ) + ( n – r ) ( n – r + ) = ( r + ) ( n – r + ) or r + r + n – nr + n – nr + r – r = ( nr – r + r + n –

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