Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If $3 \times{ }^5 \mathrm{C}_0+8 \times{ }^5 \mathrm{C}_1+13 \times{ }^5 \mathrm{C}_2+18 \times{ }^5 \mathrm{C}_3+23 \times{ }^5 \mathrm{C}_4+$ $28 \times{ }^5 \mathrm{C}_5=\mathrm{k} \times 2^4$, then $\mathrm{k}=$
MathematicsBinomial TheoremAP EAMCETAP EAMCET 2023 (16 May Shift 1)
Options:
  • A 33
  • B 37
  • C 31
  • D 30
Solution:
2534 Upvotes Verified Answer
The correct answer is: 31
$\begin{aligned} 3 \times{ }^5 C_0+8 \times{ }^5 C_1 & +13 \times{ }^5 C_2+18 \times{ }^5 C_3 \\ & +23 \times{ }^5 C_4+\ldots .+28 \times{ }^5 C_5=k \times 2^4\end{aligned}$
$\begin{aligned}=3 \times{ }^5 C_0+(5+3)^5 C_1+ & (10+3)^5 C_2+(15+3)^5 C_3 \\ & +(20+3)^5 C_4+\ldots .+(25+3)^5 C_5\end{aligned}$
$\begin{aligned}=3 \times\left({ }^5 C_6+{ }^5 C_1\right. & \left.+{ }^5 C_2+{ }^5 C_3+{ }^5 C_4+{ }^5 C_5\right) \\ & +5\left({ }^5 C_1+2{ }^5 C_2+3{ }^5 C_3+4{ }^5 C_4+5^5 C_5\right)\end{aligned}$
$=3 \times 2^5+5\left(5+2 \times \frac{5 \times 4}{2}+3 \times \frac{5 \times 4}{2}+4 \times 5+5\right)$
$\begin{aligned} & =3 \times 2^5+5 \times 80 \\ & =3 \times 2 \times 2^4+25 \times 2^4 \\ & =(6+25) \times 2^4=31 \times 2^4\end{aligned}$
$\begin{array}{r}\therefore \quad 3 \times{ }^5 C_0+8 \times{ }^5 C_1+13 \times{ }^5 C_2+18 \times{ }^5 C_3 \\ +23 \times{ }^5 C_4+28 \times{ }^5 C_5=31 \times 2^4\end{array}$
$\begin{aligned} & \Rightarrow \quad k \times 2^4=31 \times 2^4 \\ & \Rightarrow \quad k=31\end{aligned}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.