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Show that 3C_0-8C_1 + 13C_2 - 18C_3 + ..... + (n+1)^(th) term = 0 |
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Answer» Solution :`3C_0-8C_1 + 13C_2 - 18C_3 + ..... + (N+1)^TH` term = `3C_0 - (5+3)C_1 + (10+3)C_2 - (15+3)C_3` + .... = `3(C_0 - C_1 + C_2 ....) + 5(-C_1 + 2C_2 - 3C_3 ....) But `(1+X)^n` = `C_0 - C_1x + C_2x^2 - C_3x^3 + ... + (-1)^n x^nC_n ....(1) putting x = 1 we get `C_0 - C_1 + C_2 .... + (-1) "^nC_n` = 0 and differentiating (1) we get `n(1-x)^(n-1) (-1) = `-C_1 + 2C_2 x - 3C_3x^2 + .... putting x = 1 we get `-C_1 + 2C_2 - 3C_3 + .... = 0 THEREFORE `3C_0 - 8C_1 + 13C_2 ..... (n+1)` terms = 0 |
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