Let (1+2 x)20=a0+a1 x +a2 x2+ ldots +a20 x20. Then, 3 a0+2 a1+3 a2+2 a3+3 a4+2 a5+ ldots+2 a19+3 a20 equals

Question

Let (1+2 x)20=a0+a1 x +a2 x2+ ldots +a20 x20. Then, 3 a0+2 a1+3 a2+2 a3+3 a4+2 a5+ ldots+2 a19+3 a20 equals (A) (5 ⋅ 320-3/2) (B) (5 ⋅ 320+3/2) (C) (5 ⋅ 320+1/2) (D) (5 ⋅ 320-1/2)

Tardigrade Admin · Accepted Answer

(1+2 x)20=a0+a1 x+a2 x2+ ldots +a20 x20 Put x=1 320=a0+a1+a2+ ldots+a20 ...(i) Put x=-1, 1=a0-a1+a2-a3+ ldots+a20 ...(ii) On adding Eqs. (i) and (ii), we get (320+1/2)=a0+a2+a4+ ldots+a20 On subtracting Eq. (ii) from Eq. (i), we get (320-1/2)=a1+a3+a5+ ldots+a19 Now, we have 3 a0+2 a1+3 a2+2 a3+ ldots+2 a19+3 a20 =3(a0+a2+a4+ ldots+a20)+2(a1+a3+ ldots +a19) = 3((320+1/2))+2((320-1/2)) = (5 ⋅ 320+1/2)

Q. Let $(1 + 2 x)^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{20} x^{20}$ . Then, $3 a_{0} + 2 a_{1} + 3 a_{2} + 2 a_{3} + 3 a_{4} + 2 a_{5} + \dots + 2 a_{19} + 3 a_{20}$ equals

$\frac{5 \cdot 3 ^{20} - 3}{2}$

$\frac{5 \cdot 3 ^{20} + 3}{2}$

$\frac{5 \cdot 3 ^{20} + 1}{2}$

$\frac{5 \cdot 3 ^{20} - 1}{2}$

Q. Let (1+2x)20=a0​+a1​x+a2​x2+…+a20​x20. Then, 3a0​+2a1​+3a2​+2a3​+3a4​+2a5​+…+2a19​+3a20​ equals

25⋅320−3​

25⋅320+3​

25⋅320+1​

25⋅320−1​

Q. Let $(1 + 2 x)^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{20} x^{20}$ . Then, $3 a_{0} + 2 a_{1} + 3 a_{2} + 2 a_{3} + 3 a_{4} + 2 a_{5} + \dots + 2 a_{19} + 3 a_{20}$ equals

$\frac{5 \cdot 3 ^{20} - 3}{2}$

$\frac{5 \cdot 3 ^{20} + 3}{2}$

$\frac{5 \cdot 3 ^{20} + 1}{2}$

$\frac{5 \cdot 3 ^{20} - 1}{2}$