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Let `T_r` be the rth term of an A.P. whose first term is -1/2 and common difference is 1, then `sum_(r=1)^n sqrt(1+ T_r T_(r+1) T_(r+2) T_(r+3))` |
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Answer» Here, ` a= -1/2 and d = 1` `:. T_r = a+(r-1)d = -1/2+(r-1)1 = r-3/2` `:. T_rT_(r+1)T_(r+2)T_(r+3) = (r-3/2)(r-1/2)(r+1/2)(r+3/2)` `=> T_rT_(r+1)T_(r+2)T_(r+3) = (r-3/2)(r+3/2)(r-1/2)(r+1/2)` `=> T_rT_(r+1)T_(r+2)T_(r+3) = (r^2-9/4)(r^2-1/4)` `=> T_rT_(r+1)T_(r+2)T_(r+3) = r^4-10/4r^2+9/16` `:. 1+T_rT_(r+1)T_(r+2)T_(r+3) = r^4-10/4r^2+25/16 = (r^2-5/4)^2` `:. sum_(r=1)^n sqrt(1+T_rT_(r+1)T_(r+2)T_(r+3) )= sum_(r=1)^n (r^2-5/4)` `= sum_(r=1)^n r^2 - sum_(r=1)^n 5/4` `= (n(n+1)(2n+1))/6-5/4n` So, option `A` is the correct answer. |
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