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)/ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

115

No worries! We‘ve got your back. Try BYJU‘S free classes today!

120

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

135

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

Suggest Corrections

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}

Q. If

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

then

a_{1}

equals

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

, then

Join BYJU'S Learning Program

(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

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