Let a sequence be defined by a1- 0 and an-1- an-4n-3 for all n- 1 n- NThe value of ak in terms of k is k- N

The correct option is A (k 1)(2k+3) a _ n+1 = a _ n +4n+3 a _ 2 = a _ 1 +4.1+3=7=( 2 1 ) ( 2 · 2+3 ) a _ 3 = a _ 2 +4.2+3=18=( 3 1 ) ( 3 · ...

You visited us 68 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

Let a sequenc...

Question

Let a sequence be defined by $a_{1} = 0$ and $a_{n + 1} = a_{n} + 4 n + 3$ for all $n \geq 1 (n ϵ N)$

The value of $a_{k}$ in terms of k is $(k \in N)$

(k - 1) (2 k + 3)

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

(k - 1) (3 k + 1)

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

(k - 1) (k + 3)

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

(k - 1) (2 k + 5)

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

Open in App

Solution

The correct option is A $(k - 1) (2 k + 3)$
$a_{n + 1} = a_{n} + 4 n + 3$

$a_{2} = a_{1} + 4.1 + 3 = 7 = (2 - 1) (2 \cdot 2 + 3)$

$a_{3} = a_{2} + 4.2 + 3 = 18 = (3 - 1) (3 \cdot 2 + 3)$

$a_{k} = (k - 1) (2 k + 3)$

Suggest Corrections

Similar questions

A sequence $a_{1}, a_{2}, a_{3}, . . . . . . .$ is defined by letting $a_{1} = 3$ and $a_{k} = 7$ $a_{k - 1}$ for all natural numbers $k \geq 2$ . Show that $a_{n} = {3.7}^{n - 1}$ for all $n ϵ N$ .

A sequence x_{1}, x_{2}, x_{3}, . . . is defined by letting x_{1} = 2 and x_{k} = \frac{x_{k - 1}}{k} for all natural numbers k, k \geq 2 . Show that x_{n} = \frac{2}{n!} for all n \in N .

[NCERT EXEMPLAR]

A sequence $x_{1}, x_{2}, x_{3}, . . . . . .$ is defined by letting $x_{1} = 2$ and $x_{k} = \frac{x_{k - 1}}{n}$ for all natural numbers k, $k \geq 2$ . Show that $x_{n} = \frac{2}{n!}$ for all $n ϵ N$ .

A sequence (x_0, x_1, x_2, x_3, ......) is defined by letting $x_{0} = 5$ and $x_{k} = 4 + x_{k - 1}$ for all natural numbers k. Show that xn = 5 + 4n for all n $ϵ$ N using mathematical induction.

Q. Let

C_{k} =

^{n} C_{k}

for

0 \leq k \leq n

and

A_{k} = [\begin{matrix} C_{k - 1}^{2} & 0 0 & C_{k}^{2} \end{matrix}]

for

k \geq 1

, and

A_{1} + A_{2} + . . . + A_{n} = [\begin{matrix} k_{1} & 0 0 & k_{2} \end{matrix}]

, then

Join BYJU'S Learning Program

Let a sequence be defined by a1=0 and an+1=an+4n+3 for all n≥1(nϵN)The value of ak in terms of k is (k∈N)

The correct option is A (k−1)(2k+3)an+1=an+4n+3a2=a1+4.1+3=7=(2−1)(2⋅2+3)a3=a2+4.2+3=18=(3−1)(3⋅2+3)ak=(k−1)(2k+3)

Let a sequence be defined by $a_{1} = 0$ and $a_{n + 1} = a_{n} + 4 n + 3$ for all $n \geq 1 (n ϵ N)$

The value of $a_{k}$ in terms of k is $(k \in N)$

The correct option is A $(k - 1) (2 k + 3)$
$a_{n + 1} = a_{n} + 4 n + 3$

$a_{2} = a_{1} + 4.1 + 3 = 7 = (2 - 1) (2 \cdot 2 + 3)$

$a_{3} = a_{2} + 4.2 + 3 = 18 = (3 - 1) (3 \cdot 2 + 3)$

$a_{k} = (k - 1) (2 k + 3)$