Sum up to $3 n$ terms $1, 3, - 5, 7, 9, - 11, 13, 15, - 17...$ is

n (3 n - 4)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3 n (n - 4)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n (3 n + 4)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(3 n - 4)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $n (3 n - 4)$
$S_{n} = (1 + 3 - 5) + (7 + 9 - 11) + (13 + 15 - 17) + . . . .$ up to $3 n$ terms
$\Rightarrow S_{n} = - 1 + 5 + 11 + . . . .$ up to $n$ terms
Since, $- 1, 5, 11, . . .$ are in A.P with first term $- 1$ and common difference $6$
Therefore, $S_{n} = \frac{n [2 a + (n - 1) d]}{2} = \frac{n [- 2 + (n - 1) 6]}{2} = n (3 n - 4)$
Ans: $A$ .

Suggest Corrections

Similar questions

Prove using mathematical induction that for all $n \geq 1$

1+4+7+..+(3n-2)= $\frac{n (3 n - 1)}{2}$

1, z_{1}, z_{2}, . . . z_{n - 1}

\frac{1}{3 - z_{1}} + \frac{1}{3 - z_{2}} + . . . \frac{1}{3 - z_{n - 1}}

I f n \sum i = 1 i = \frac{n (n + 1)}{2}, t h e n n \sum i = 1 (3 i - 2) =

n \sum i - 1 i = \frac{n (n + 1)}{2}, t h e n n \sum i - 1 (3 i - 2) =

Prove using mathematical induction that for all $n \geq 1$

1+4+7+..+(3n-2)= $\frac{n (3 n - 1)}{2}$

Join BYJU'S Learning Program