The value of -15C1-2-15C2-3-15C3-15-15C15-14C1-14C3-14C5-14C11 is -

S_1= ^15C_1+2×^15C_2 3×^15C_3+... 15×^15C_15 =∑_r=1^15( 1)^r× r×^15C_r=15∑_r=1^15( 1)^r^14C_r 1 nbsp; =15( ^14C_0+^14C_1 ^14C_2+⋯ ^14C_14)=15(0)=0 S_2=^14 ...

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Byju's Answer

Standard XI

Mathematics

Sum of coefficients of all terms

The value of ...

Question

The value of $-^{15} C_{1} + 2 \times^{15} C_{2} - 3 \times^{15} C_{3} + . . . - 15 \times^{15} C_{15} +^{14} C_{1} +^{14} C_{3} +^{14} C_{5} + . . . +^{14} C_{11}$ is :

2^{14}

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2^{13} - 13

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2^{16} - 1

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2^{13} - 14

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Solution

The correct option is D $2^{13} - 14$
$S_{1} = -^{15} C_{1} + 2 \times^{15} C_{2} - 3 \times^{15} C_{3} + . . . - 15 \times^{15} C_{15}$
$= 15 \sum r = 1 (- 1)^{r} \times r \times^{15} C_{r} = 15 15 \sum r = 1 (- 1)^{r}^{14} C_{r - 1} = 15 (-^{14} C_{0} +^{14} C_{1} -^{14} C_{2} + \dots -^{14} C_{14}) = 15 (0) = 0 S_{2} =^{14} C_{1} +^{14} C_{3} +^{14} C_{5} + \dots +^{14} C_{11} = (^{14} C_{1} +^{14} C_{3} +^{14} C_{5} + \dots +^{14} C_{11} +^{14} C_{13}) -^{14} C_{13} = 2^{13} - 14 ∴ S_{1} + S_{2} = 2^{13} - 14$

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Similar questions

^{14} C_{1} +^{14} C_{2} +^{14} C_{3} + . . . . +^{14} C_{14} =

\frac{^{15} C_{1}}{^{15} C_{0}} + 2. \frac{^{15} C_{2}}{^{15} C_{1}} + 3. \frac{^{15} C_{3}}{^{15} C_{2}} + \dots + 15. \frac{^{15} C_{15}}{^{15} C_{14}} =

Q. The value of

^{15} C_{0} +^{15} C_{1} +^{15} C_{2} + \dots +^{15} C_{7}

The Value of $^{14} C_{2} +^{14} C_{3} +^{14} C_{4} + . . . . . +^{14} C_{14}$ is

Find the value of $^{15} C_{2}$ + 2 $^{15} C_{3}$ + 3 $^{15} C_{4}$ . . . . . 14 $^{15} C_{15}$

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The value of −15C1+2×15C2−3×15C3+...−15×15C15+14C1+14C3+14C5+...+14C11 is :

The correct option is D 213−14S1=−15C1+2×15C2−3×15C3+...−15×15C15 =15∑r=1(−1)r×r×15Cr=1515∑r=1(−1)r14Cr−1=15(−14C0+14C1−14C2+⋯−14C14)=15(0)=0 S2=14C1+14C3+14C5+⋯+14C11=(14C1+14C3+14C5+⋯+14C11+14C13)−14C13=213−14∴S1+S2=213−14

The value of $-^{15} C_{1} + 2 \times^{15} C_{2} - 3 \times^{15} C_{3} + . . . - 15 \times^{15} C_{15} +^{14} C_{1} +^{14} C_{3} +^{14} C_{5} + . . . +^{14} C_{11}$ is :