The value of 2c-0-2-2-2C-1-2-3-3C-2-2-4-4C-3-2-11-11C-10-is

The correct option is A (1+x)^10=^10C_0+^10C_2x^2+....+^10C_0×10 Integrating w.r.t. x between 0 and 2, we get ((1+x)^10/11 )^2_0= | ( ^10C_0+^10C_1+x^ ...

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Byju's Answer

Standard XI

Mathematics

Integration to solve modified sum of binomial coefficients

The value of ...

Question

The value of $2 c_{0} + \frac{2^{2}}{2} C_{1} + \frac{2^{3}}{3} C_{2} + \frac{2^{4}}{4} C_{3} + . . . . + \frac{2^{11}}{11} C_{10} i s$

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Solution

The correct option is A
$(1 + x)^{10} =^{10} C_{0} +^{10} C_{2} x^{2} + . . . . +^{10} C_{0} \times 10 I n t e g r a t i n g w . r . t . x b e t w e e n 0 a n d 2, w e g e t {(\frac{(1 + x)^{10}}{11})}_{0}^{2} = {∣ ∣ (^{10} C_{0} +^{10} C_{1} + \frac{x^{2}}{2} +^{10} C_{1} \frac{x^{2}}{2} +^{10} C_{2} \frac{x^{3}}{3} + . . .^{10} C_{10} + \frac{x^{11}}{11}) ∣ ∣}_{0}^{2} \Rightarrow \frac{3^{11}}{11} - \frac{1}{11} = (^{10} C_{0} 2 + 10_{C_{1}} \frac{2^{2}}{2} + 10_{C_{2}} \frac{2_{3}}{3} + . . . + 10_{C_{10}} \frac{2^{11}}{11}) - 0 \Rightarrow^{2} C_{0} + \frac{2^{2}}{2} c_{1} + \frac{2^{3}}{3} c_{2} + . . . + \frac{2^{11}}{11} C_{10} = \frac{3^{11} - 1}{11}$

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Similar questions

Q. The value of

2 c_{0} + \frac{2^{2}}{2} C_{1} + \frac{2^{3}}{3} C_{2} + \frac{2^{4}}{4} C_{3} + . . . . + \frac{2^{11}}{11} C_{10} i s

Q. The value of

2 c_{0} + \frac{2^{2}}{2} C_{1} + \frac{2^{3}}{3} C_{2} + \frac{2^{4}}{4} C_{3} + . . . . + \frac{2^{11}}{11} C_{10} i s

Q. Find the sum

2 C_{o} + \frac{2^{2}}{2} C_{1} + \frac{2^{3}}{3} C_{2} + \frac{2^{4}}{4} C_{3} + . . . + \frac{2^{11}}{11} C_{10}

Q. In the usual notation prove that

2 C_{0} + \frac{2^{2}}{2} C_{1} + \frac{2^{3}}{3} C_{2} + . . . . . . + \frac{2^{11}}{11} C_{10} = \frac{3^{11} - 1}{11}

Q. If C

_{r}

stands for

^{n} C_{r}

, then the value of

2 C_{0} - \frac{2^{2}}{2} C_{1} + \frac{2^{3}}{3} C_{2} + . . . . . + (- 1)^{n + 1} \frac{2^{n + 1}}{(n + 1)}

is equal to

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The value of 2c0+222C1+233C2+244C3+....+21111C10 is

The correct option is A (1+x)10=10C0+10C2x2+....+10C0×10Integrating w.r.t. x between 0 and 2, we get((1+x)1011)20=∣∣(10C0+10C1+x22+10C1x22+10C2x33+...10C10+x1111)∣∣20⇒31111−111=(10C0 2+10C1222+10C2233+...+10C1021111)−0⇒2C0+222c1+233c2+...+21111C10=311−111

The value of $2 c_{0} + \frac{2^{2}}{2} C_{1} + \frac{2^{3}}{3} C_{2} + \frac{2^{4}}{4} C_{3} + . . . . + \frac{2^{11}}{11} C_{10} i s$