यदि n,1 से बड़ा एक पूर्णांक है,तो a - ^nC_1(a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $S = \sum_{k=0}^{n} (-1)^k {^nC_k} (a - k)$.
This can be written as $S = a \sum_{k=0}^{n} (-1)^k {^nC_k} - \sum_{k=0}^{n} (-1)^k k {^nC_k}$.

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Let $S = \sum_{k=0}^{n} (-1)^k {^nC_k} (a - k)$.
This can be written as $S = a \sum_{k=0}^{n} (-1)^k {^nC_k} - \sum_{k=0}^{n} (-1)^k k {^nC_k}$.
We know that $\sum_{k=0}^{n} (-1)^k {^nC_k} = (1 - 1)^n = 0$ for $n \ge 1$.
Also, $\sum_{k=0}^{n} (-1)^k k {^nC_k} = 0$ for $n > 1$ because $k {^nC_k} = n {^{n-1}C_{k-1}}$, and the sum becomes $n \sum_{k=1}^{n} (-1)^k {^{n-1}C_{k-1}} = n(1 - 1)^{n-1} = 0$.
Thus, $S = a(0) - 0 = 0$.

यदि $n$,$1$ से बड़ा एक पूर्णांक है,तो $a - ^nC_1(a - 1) + ^nC_2(a - 2) + \dots + (-1)^n(a - n) = $

Similar Questions

$\sum\limits_{r = 0}^m {^{n + r}{C_n} = } $

यदि $C_0, C_1, C_2, \ldots, C_{10}$ द्विपद गुणांक $(1+x)^{10}$ के विस्तार में हैं,तो $C_0 C_6+C_1 C_7+C_2 C_8+C_3 C_9+C_4 C_{10}=$

यदि $n$ एक धनात्मक पूर्णांक है, तो $(2n+1) ^nC_0 + (2n-1) ^nC_1 + (2n-3) ^nC_2 + \ldots + 1 \cdot ^nC_n$ का मान क्या है?

$\binom{47}{4} + \sum_{r=1}^5 \binom{52-r}{3} = \dots$

मान लीजिए कि $a_0, a_1, a_2, \ldots, a_n \in \mathbb{R}$ एक समांतर श्रेणी में हैं और $C_0, C_1, C_2, \ldots, C_n$ द्विपद गुणांक हैं। तो $\sum_{k=0}^n a_k \cdot C_k =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module