Tính B = 1A2^2 + 1A3^2 + ... + 1An^2 biết Cn^1 + 2Cn^2Cn^1 + ... + nCn^nCn^n - 1 = 45.
Câu hỏi
Nhận biếtTính \(B = \frac{1}{{A_2^2}} + \frac{1}{{A_3^2}} + ... + \frac{1}{{A_n^2}}\), biết \(C_n^1 + 2\frac{{C_n^2}}{{C_n^1}} + ... + n\frac{{C_n^n}}{{C_n^{n - 1}}} = 45\).
Đáp án đúng: A
Lời giải của Tự Học 365
Giải chi tiết:
Ta có: \(C_n^1 = n;\;\;\;2\frac{{C_n^2}}{{C_n^1}} = 2.\frac{{\frac{{n!}}{{2!.(n - 2)!}}}}{{\frac{{n!}}{{1!.(n - 1)!}}}} = 2.\frac{{n\left( {n - 1} \right)}}{{2.n}} = n - 1;......;\;n\frac{{C_n^n}}{{C_n^{n - 1}}} = n.\frac{1}{{\frac{{n!}}{{1!.(n - 1)!}}}} = \frac{n}{n} = 1.\)
\(\begin{array}{l} \Rightarrow C_n^1 + 2\frac{{C_n^2}}{{C_n^1}} + ... + n\frac{{C_n^n}}{{C_n^{n - 1}}} = 45 \Leftrightarrow 1 + 2 + 3 + .. + n = 45\\ \Leftrightarrow \frac{{n(n + 1)}}{2} = 45 \Leftrightarrow {n^2} + n - 90 = 0 \Leftrightarrow \left[ \begin{array}{l}n = 9\;\;\left( {tm} \right)\\n = -10\;\;\left( {ktm} \right)\end{array} \right.\\B = \frac{1}{{A_2^2}} + \frac{1}{{A_3^2}} + ... + \frac{1}{{A_n^2}} = \frac{1}{{\frac{{2!}}{{\left( {2 - 2} \right)!}}}} + \frac{1}{{\frac{{3!}}{{\left( {3 - 2} \right)!}}}} + ...... + \frac{1}{{\frac{{n!}}{{\left( {n - 2} \right)!}}}}\\ = \frac{{0!}}{{2!}} + \frac{{1!}}{{3!}} + \frac{{2!}}{{4!}} + .... + \frac{{(n - 2)!}}{{n!}} = \frac{1}{2} + \frac{1}{6} + \frac{1}{{24}} + ..... + \frac{1}{{n\left( {n - 1} \right)}}\\ = \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{n.(n - 1)}}\\ = 1 - \frac{1}{2} + \frac{1}{2} - .... + \frac{1}{{n - 1}} - \frac{1}{n}\\ = 1 - \frac{1}{n} = 1 - \frac{1}{{9}} = \frac{8}{{9}}.\end{array}\)
Chọn A
