TS EAMCET · Maths · Binomial Theorem
\(\sum_{r=0}^{10}{ }^{40-r} C_5\) is equal to
- A \({ }^{41} C_5-{ }^{30} C_5\)
- B \({ }^{41} C_6-{ }^{30} C_6\)
- C \({ }^{41} \mathrm{C}_5+{ }^{30} \mathrm{C}_5\)
- D \({ }^{41} C_6\)
Answer & Solution
Correct Answer
(B) \({ }^{41} C_6-{ }^{30} C_6\)
Step-by-step Solution
Detailed explanation
\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}…
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