Let n in N and [x] denote the greatest intege | vedclass

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(D) The general term of the series is $T_{r+1} = (2r+1) \cdot {}^{n}C_{r}$ for $r = 0, 1, 2, \ldots, n$.The sum $S$ is given by $S = \sum_{r=0}^{n} (2

Vedclass · Accepted Answer

$98$

Vedclass · Answer

(D) The general term of the series is $T_{r+1} = (2r+1) \cdot {}^{n}C_{r}$ for $r = 0, 1, 2, \ldots, n$.The sum $S$ is given by $S = \sum_{r=0}^{n} (2r+1) \cdot {}^{n}C_{r}$.$S = 2 \sum_{r=0}^{n} r \cdot {}^{n}C_{r} + \sum_{r=0}^{n} {}^{n}C_{r}$.Using the identities $\sum r \cdot {}^{n}C_{r} = n \cdot 2^{n-1}$ and $\sum {}^{n}C_{r} = 2^{n}$,we get:$S = 2(n \cdot 2^{n-1}) + 2^{n} = n \cdot 2^{n} + 2^{n} = 2^{n}(n+1)$.Given $S = 2^{100} \cdot 101$,we have $2^{n}(n+1) = 2^{100} \cdot 101$,which implies $n = 100$.Now,calculate $2\left[\frac{n-1}{2}\right] = 2\left[\frac{100-1}{2}\right] = 2\left[\frac{99}{2}\right] = 2[49.5] = 2 \cdot 49 = 98$.

Let $n \in N$ and $[x]$ denote the greatest integer less than or equal to $x$. If the sum of $(n+1)$ terms ${}^{n}C_{0}, 3 \cdot {}^{n}C_{1}, 5 \cdot {}^{n}C_{2}, 7 \cdot {}^{n}C_{3}, \ldots$ is equal to $2^{100} \cdot 101$,then $2\left[\frac{n-1}{2}\right]$ is equal to $....$

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