Question 286 of 347MediumMCQ
If \[{S_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}}} \,\,and\,\,{t_n} = \sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}},\,} \]then \[{{{t_{ n}}} \over {{S_n}}}\] is equal to
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Official solution
\[{S_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}}}\]
=\[{1 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + .... + {1 \over {{}^n{C_n}}}\]
\[{t_n} = \sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}}}\]
= \[{0 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + {2 \over {{}^n{C_2}}}.... + {n \over {{}^n{C_n}}}\].........(1)
We can write tn by rearranging like this,
tn = \[{n \over {{}^n{C_n}}} + {{n - 1} \over {{}^n{C_{n - 1}}}} + ... + {1 \over {{}^n{C_1}}} + {0 \over {{}^n{C_0}}}\]
= \[{n \over {{}^n{C_0}}} + {{n - 1} \over {{}^n{C_1}}} + ... + {1 \over {{}^n{C_{n - 1}}}} + {0 \over {{}^n{C_n}}}\].........(2)
[as nC0 = nCn, nC1 = nCn−1......]
By adding (1) and (2) we get,
2tn = \[{n \over {{}^n{C_0}}} + {n \over {{}^n{C_1}}} + ... + {n \over {{}^n{C_{n - 1}}}} + {n \over {{}^n{C_n}}}\]
= \[n\left[ {{1 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + ... + {1 \over {{}^n{C_{n - 1}}}} + {1 \over {{}^n{C_n}}}} \right]\]
= nSn
\[\therefore\] \[{{{t_n}} \over {{S_n}}} = {n \over 2}\] Answer the question to unlock