Question

If (1 + x)¹³ = a₀ + a₁x + . . . . .  + a₁₃ x¹³ then, find the value of \(\rm \sum _{r=1}^{13}r\frac{a_r}{a_{r-1}}\)
1. 104
2. 91
3. 169
4. 182

Answer 1

Correct Answer - Option 2 : 91

Concept:

(1 + x)n =  nC0 + nC1 x + nC2 x2 +...+ nCn xn

ⁿC₁ = n

ⁿC_n = 1

 

Calculation:

Here, (1 + x)13 = a0 + a1x + . . . . .  + a13 x13                         ....(1)

\(\rm (1+x)^{13}=1+^{13}C_1(x)+^{13}C_2(x^2)+....+^{13}C_{13}(x^{13})\)         ....(2)

From (1) and (2), we get

\(\rm a_0=1, a_1=^{13}C_1, a_2=^{13}C_2,.......,a_{13}=^{13}C_{13}\)

Now,

 \(\rm \sum _{r=1}^{13}r\frac{a_r}{a_{r-1}}=(1\times \frac{a_1}{a_0})+(2\times \frac{a_2}{a_1})+......+(13\times \frac{a_{13}}{a_{12}})\)

\(\rm =(1\times \frac{13C_1}{1})+(2\times \frac{13C_2}{13C_1})+......+(13\times \frac{13C_{13}}{13C_{12}})\)

\(\rm =(13)+(2\times \frac{13\times 12}{2\times13})+......+(13\times \frac{1}{13})\)

= 13 + 12 + 11 + .......+ 1

= 91

Hence, option (2) is correct.