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$\sum_{r=0}^{10}{ }^{40-r} C_5$ is equal to
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1419 Upvotes
Verified Answer
The correct answer is:
${ }^{41} C_6-{ }^{30} C_6$
$$
\begin{aligned}
& \text { } \sum_{r=0}^{10}{ }^{40-r} C_5={ }^{30} C_5+{ }^{31} C_5+{ }^{32} C_5+{ }^{33} C_5 \\
& +\ldots+{ }^{40} C_5 \\
& \because{ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r \\
& \Rightarrow \quad{ }^n C_{r-1}={ }^{n+1} C_r-{ }^n C_r \\
& \therefore \quad{ }^{30} C_5={ }^{31} C_6-{ }^{30} C_6 \\
& { }^{31} C_5={ }^{32} C_6-{ }^{31} C_6 \\
&
\end{aligned}
$$
On adding, we get
$$
\sum_{r=0}^{10}{ }^{40-r} C_5={ }^{41} C_6-{ }^{30} C_6
$$
\begin{aligned}
& \text { } \sum_{r=0}^{10}{ }^{40-r} C_5={ }^{30} C_5+{ }^{31} C_5+{ }^{32} C_5+{ }^{33} C_5 \\
& +\ldots+{ }^{40} C_5 \\
& \because{ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r \\
& \Rightarrow \quad{ }^n C_{r-1}={ }^{n+1} C_r-{ }^n C_r \\
& \therefore \quad{ }^{30} C_5={ }^{31} C_6-{ }^{30} C_6 \\
& { }^{31} C_5={ }^{32} C_6-{ }^{31} C_6 \\
&
\end{aligned}
$$
On adding, we get
$$
\sum_{r=0}^{10}{ }^{40-r} C_5={ }^{41} C_6-{ }^{30} C_6
$$
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