The sum to n terms of the series 1+54n+14n−3+94n+14n−32+134n+14n−33+…. is

Lets =1+5x+9x2+……+(4n−3)xn−1⇒xs=0+x+5x2+…………+(4n−3)xn(1−x)s=1+4x+4x2………+4xn−1−(4n−3)xn(1−x)s=41−xn1−x−3−(4n−3)xn,x=4n+14n−31−4n+14n−3nS=41−4n+14n−3n1−4n+14n−3−3−(4n−3)4n+14n−3n−44n−3S=−(4n−3)1−4n−14n−3n−3(4n−3)4n+14n−3n4S=(4n−3)2+3(4n−3)4S=(4n−3)(4n−3+3),S=n(4n−3)

The sum to n terms of the series 1+54n+14n−3+94n+14n−32+134n+14n−33+…. is

Lets =1+5x+9x2+……+(4n−3)xn−1⇒xs=0+x+5x2+…………+(4n−3)xn(1−x)s=1+4x+4x2………+4xn−1−(4n−3)xn(1−x)s=41−xn1−x−3−(4n−3)xn,x=4n+14n−31−4n+14n−3nS=41−4n+14n−3n1−4n+14n−3−3−(4n−3)4n+14n−3n−44n−3S=−(4n−3)1−4n−14n−3n−3(4n−3)4n+14n−3n4S=(4n−3)2+3(4n−3)4S=(4n−3)(4n−3+3),S=n(4n−3)

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The sum to n terms of the series $1 + 5 (\frac{4 n + 1}{4 n - 3}) + 9 {(\frac{4 n + 1}{4 n - 3})}^{2} + 13 {(\frac{4 n + 1}{4 n - 3})}^{3} + \dots .$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$(n + 1) (4 n - 3)$

$n (4 n - 3)$

$(n + 1) (4 n)$

$n (4 n + 3)$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} Lets = 1 + 5 x + 9 x^{2} + \dots \dots + (4 n - 3) x^{n - 1} \\ \Rightarrow xs = 0 + x + 5 x^{2} + \dots \dots \dots \dots + (4 n - 3) x^{n} \\ (1 - x) s = 1 + 4 x + 4 x^{2} \dots \dots \dots + 4 x^{n - 1} - (4 n - 3) x^{n} \\ (1 - x) s = 4 (\frac{1 - x^{n}}{1 - x}) - 3 - (4 n - 3) x^{n}, x = \frac{4 n + 1}{4 n - 3} \\ {(1 - (\frac{4 n + 1}{4 n - 3}))}^{n} S = \frac{4 (1 - {(\frac{4 n + 1}{4 n - 3})}^{n})}{1 - (\frac{4 n + 1}{4 n - 3})} \\ - 3 - (4 n - 3) {(\frac{4 n + 1}{4 n - 3})}^{n} \frac{- 4}{4 n - 3} S = - (4 n - 3) \\ (1 - {(\frac{4 n - 1}{4 n - 3})}^{n}) - 3 (4 n - 3) {(\frac{4 n + 1}{4 n - 3})}^{n} \end{array}$
$\begin{array}{l} 4 S = (4 n - 3)^{2} + 3 (4 n - 3) \\ 4 S = (4 n - 3) (4 n - 3 + 3), S = n (4 n - 3) \end{array}$