Sum upto n terms 1.3 + 3.5 + 5.7 + …… + Tn = | MATH - ARITHMETIC PROGRESSION | SolveeIt

Sum upto n terms 1.3 + 3.5 + 5.7 + …… + T_n =

$\frac { 1 } { 3 }$ n (4n² + 6n – 1)

$\frac { 1 } { 3 }$ n(4n² + 3n + 3)

$\frac { 1 } { 6 }$ n (4n² + 6n – 1)

None

Answer

$\frac { 1 } { 3 }$ n (4n² + 6n – 1)

Explanation

T_n = (2n – 1).(2n + 1) = 4n² –1

S_n = S (4n² – 1)= $\frac { 4 \mathrm { n } ( \mathrm { n } + 1 ) ( 2 \mathrm { n } + 1 ) } { 6 } - \mathrm { n }$

= (4n² + 6n – 1)

Question: Sum upto n terms 1.3 + 3.5 + 5.7 + …… + T<sub>n</sub> =...