- Home
- Standard 11
- Mathematics
यदि $n \geq 2$ एक धनात्मक पूर्णाक है, तो श्रेणी ${ }^{n+1} C _{2}+2\left({ }^{2} C _{2}+{ }^{3} C _{2}+{ }^{4} C _{2}+\ldots+{ }^{2} C _{2}\right)$ का योग है
$\frac{ n ( n -1)(2 n +1)}{6}$
$\frac{ n ( n +1)(2 n +1)}{6}$
$\frac{ n (2 n +1)(3 n +1)}{6}$
$\frac{ n ( n +1)^{2}( n +2)}{12}$
Solution
${ }^{n+1} C_{2}+2\left({ }^{2} C_{2}+{ }^{3} C_{2}+{ }^{4} C_{2}+\ldots \ldots .+{ }^{n} C_{2}\right)$
${ }^{n+1} C_{2}+2\left({ }^{3} C_{3}+{ }^{3} C_{2}+{ }^{4} C_{2}+\ldots \ldots .+{ }^{n} C_{2}\right)$
$\left\{\right.$ use $\left.{ }^{n} C_{r+1}+{ }^{n} C_{r}={ }^{n+1} C_{r}\right\}$
$={ }^{n+1} C_{2}+2\left({ }^{4} C_{3}+{ }^{4} C_{2}+{ }^{5} C_{3}+\ldots \ldots .+{ }^{n} C_{2}\right)$
$={ }^{ n +1} C _{2}+2\left({ }^{5} C _{3}+{ }^{5} C _{2}+\ldots \ldots .+{ }^{ n } C _{2}\right)$
$\quad\quad\quad\quad\quad\quad\vdots \quad\quad\quad \vdots\quad\quad\quad\quad \vdots \quad\quad \vdots$
$={ }^{ n +1} C _{2}+2\left({ }^{ n } C _{3}+{ }^{ n } C _{2}\right)$
$={ }^{ n +1} C _{2}+2 \cdot{ }^{ n +1} C _{3}$
$=\frac{( n +1) n }{2}+2 \cdot \frac{( n +1)( n )( n -1)}{2.3}$
$=\frac{ n ( n +1)(2 n +1)}{6}$
