BITSAT 2017 · Mathematics · Permutation and Combination
A. 5(2n−9)5\left(2n - 9\right)5(2n−9)
We have,
∑r=110r⋅CrnCr−1n=∑r=110(n−r+1)\mathop{∑}_{r = 1}^{10}r \cdot \dfrac{Cr n}{Cr - 1 n} = \mathop{∑}_{r = 1}^{10}\left(n - r + 1\right)∑r=110r⋅Cr−1nCrn=∑r=110(n−r+1)
⇒∑r=110r⋅CrnCr−1n=∑r=110{(n+1)−r}⇒ \mathop{∑}_{r = 1}^{10}r \cdot \dfrac{Cr n}{Cr - 1 n} = \mathop{∑}_{r = 1}^{10}\left\{\left(n + 1\right) - r\right\}⇒∑r=110r⋅Cr−1nCrn=∑r=110{(n+1)−r}
⇒∑r=110r⋅CrnCr−1n=10(n+1)−∑r=110r⇒ \mathop{∑}_{r = 1}^{10}r \cdot \dfrac{Cr n}{Cr - 1 n} = 10\left(n + 1\right) - \mathop{∑}_{r = 1}^{10}r⇒∑r=110r⋅Cr−1nCrn=10(n+1)−∑r=110r
⇒∑r=110r⋅CrnCr−1n=10(n+1)−55⇒ \mathop{∑}_{r = 1}^{10}r \cdot \dfrac{Cr n}{Cr - 1 n} = 10\left(n + 1\right) - 55⇒∑r=110r⋅Cr−1nCrn=10(n+1)−55
=10n−45=5(2n−9)= 10n - 45 = 5\left(2n - 9\right)=10n−45=5(2n−9)
