Question

The sum of 19 terms of the series $S_n$= 4+10 + 26 + 66 + 162 + 386 + ...... is

• A. 19*2¹⁹ + 27
• B. 18*2¹⁹ - 27
• C. 18*2¹⁹ + 27
• D. 19*2¹⁹ - 27

Answer 1

The correct option is C

18*2¹⁹ + 27

$S_n$ = 4+10 + 26 + 66 + 162 + 386 + ......

General term = 2 + $n{2}^n$

$LetS=38+2^{\ast }{2}^2+3^{\ast }{2}^3+4^{\ast }{2}^4+\cdots \cdots +{19}^{\ast }{2}^{19}-------\left(i\right)$

$\left(\frac{S}{2}\right)=19+2^{\ast }2+3^{\ast }{2}^2+\cdots \cdots +{19}^{\ast }{2}^{18}--------\left(ii\right)$

Subtracting (ii) from (i), we get

$\left(\frac{S}{2}\right)=19+\left(2^{\ast }2\right)+{19}^{\ast }219-(2^2+2^3+2^4+2^5+2^6+..........+2^{18})$

$=19+\left(2^{\ast }2\right)+{19}^{\ast }{2}^{19}–(2^0+2^1+2^2+2^3+2^4+2^5+2^6+..........+2^{18})+3$

$=19+\left(2^{\ast }2\right)+{19}^{\ast }{2}^{19}–(2^{19}-1)+3$

$={18}^{\ast }{2}^{19}+27$