JEE Mains · Maths · STD 11 - 7. binomial theoram
\(2.{}^{20}{C_0} + 5.{}^{20}{C_1} + 8.{}^{20}{C_2} + 11.{}^{20}{C_3} + ......62.{}^{20}{C_{20}}\) is equal to
- A \({2^{23}}\)
- B \({2^{26}}\)
- C \({2^{24}}\)
- D \({2^{25}}\)
Answer & Solution
Correct Answer
(D) \({2^{25}}\)
Step-by-step Solution
Detailed explanation
\(2 \cdot^{20} \mathrm{C}_{0}+5 \cdot^{20} \mathrm{C}_{1}+8 \cdot^{20} \mathrm{C}_{2}+11 \cdot^{20} \mathrm{C}_{3}+\ldots \ldots+62 \cdot^{20} \mathrm{C}_{20} 2\) \( = \sum\limits_{r = 0}^{20} {(3r + 2){\,^{20}}} {C_r}\)…
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