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CMI_2011_TestB[S]

CMI_2011_TestB[S] - CMI Class Test-B[Q1[Total 8 marks[Sol(a...

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CMI: Class Test-B [ Sol ] [ Q1 ] [ Total: 8 marks ] (a) Prove, by mathematical induction, that 2 + 4 + 6 + ··· + 2 n = n ( n + 1 ) for all positive integers n . [ 5 ] (b) By using the identity ( k + 2 ) 3 - k 3 = 6 k 2 + 12 k + 8, or otherwise, find 2 2 + 4 2 + 6 2 + ··· + 100 2 . [ 3 ] [ Sol ] (a) For n = 1, L.H.S. = 1 = R.H.S. The statement is true for n = 1. Assume that the statement is true for some integer k 1, i.e., 2 + 4 + 6 + ··· + 2 k ) = k ( k + 1 ) For n = k + 1, 2 + 4 + 6 + ··· + 2 ( k + 1 ) = k ( k + 1 ) z }| { 2 + 4 + 6 + ··· + 2 k +( 2 k + 2 ) = k ( k + 1 ) + 2 ( k + 1 ) = ( k + 1 )[( k + 1 ) + 1 ] The statement is true for n = k + 1. Hence, by the principle of M.I., the statement is true for all positive integers n . (b) For k = 2,4,6,...,100, we have 4 3 - 2 3 = 6 ( 2 ) 2 + 12 ( 2 ) + 8 6 3 - 4 3 = 6 ( 4 ) 2 + 12 ( 4 ) + 8 8 3 - 6 3 = 6 ( 6 ) 2 + 12 ( 6 ) + 8 . . . 102 3 - 100 3 = 6 ( 100 ) 2 + 12 ( 100 ) + 8 . Adding up these 50 equations, we have 102 3 - 2 3 = 6 ( 2 2 + 4 2 + 6 2 + ··· + 100 2 ) + 12 ( 2 + 4 + 6 + ··· + 100 ) + 8 × 50 2 2 + 4 2 + 6 2 + ··· + 100 2 = 1061200 - 12 × 50 ( 51 ) - 400 6 = 171700. [ ALT ] 2 2 + 4 2 + 6 2 + ··· + 100 2 = 4 ( 1 2 + 2 2 + 3 2 + ··· + 50 2 ) = 4 50 ( 50 + 1 )[ 2 ( 50 ) + 1 ] 6 = 171700 AD: College Mathematics I [ 2011-12 ] 1
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CMI: Class Test-B [ Sol ] [ Q2 ] [ Total: 12 marks ] (a) 1 p x - 3 p x 9 is expanded in descending powers of x . (i) Write down the general term of the expansion. [ 2 ] (ii) Find the coefficient of the x - 1 / 3 term in the expansion. [ 2 ] (iii) Find the 7th term in the expansion. [
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