1- x 15- a 0- a 1 x - a 15 x 15- r -115 r a r r -1-A- 115B- 120C- 110D- 135

The correct option is B 120 ∑ ^15_r=1ra_r/a_r 1=a_1/a_0+2a_2/a_1+3a_3/a_2+……+15.a_15/a_14=C_1/C_0+2C_2/C_1+3C_3/C_2+ ……15.C_15/C_14 =15+14+13+ …… +1=15(15+1)/ ...

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Byju's Answer

Standard XII

Mathematics

Greatest Binomial Coefficients

1+ x 15= a 0+...

Question

$(1 + x)^{15} = a_{0} + a_{1} x + \dots + a_{15} x^{15} \Rightarrow \sum_{r = 1}^{15} r \frac{a_{r}}{a_{r - 1}}$ =

110

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115

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120

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135

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Solution

The correct option is C 120
$\sum_{r = 1}^{15} r \frac{a_{r}}{a_{r} - 1} = \frac{a_{1}}{a_{0}} + 2 \frac{a_{2}}{a_{1}} + 3 \frac{a_{3}}{a_{2}} + \dots \dots + 15. \frac{a_{1} 5}{a_{1} 4} = \frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots \dots 15. \frac{C_{1} 5}{C_{1} 4} = 15 + 14 + 13 + \dots \dots + 1 = \frac{15 (15 + 1)}{2}$ =120

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Similar questions

(1 + x)^{15} = a_{0} + a_{1} x + \dots \dots + a_{15} x^{15} \Rightarrow \sum_{r = 1}^{15} r \frac{a_{r}}{a_{r - 1}} =

Q. If

(1 + x)^{15} = a_{0} + a_{1} x + \dots \dots + a_{15} x^{15}

, then

\sum_{r = 1}^{15} r \frac{a_{r}}{a_{r} - 1}

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(1 + 2 x + 3 x^{2})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{20} x^{20},

then

a_{1}

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$(2 + x + x^{3})^{5} = a_{0} + a_{1} x + . . . . . + a_{15} x^{15}$ . find the value of $a_{0} + 2 a_{1} + a_{2} + 2 a_{3} . . . . + 2 a_{15}$

Q. If

(1 + x)^{2 n} = a_{0} + a_{1} x . . . . + a_{2 n} x^{2 n}

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(1+x)15=a0+a1x+…+a15x15⇒∑15r=1rarar−1=

The correct option is C 120 ∑15r=1rarar−1=a1a0+2a2a1+3a3a2+……+15.a15a14=C1C0+2C2C1+3C3C2+……15.C15C14=15+14+13+……+1=15(15+1)2=120

$(1 + x)^{15} = a_{0} + a_{1} x + \dots + a_{15} x^{15} \Rightarrow \sum_{r = 1}^{15} r \frac{a_{r}}{a_{r - 1}}$ =