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Find the sum of all numbers of the form $2 k + 1$ , where $k$ takes on integral values from $1$ to $n$ .

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n (n + 20)

n (n + 24

)

n (n + 2)

n (2 n + 2)

answer is C.

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Detailed Solution

We need to find the sum of numbers from

2 k + 1

.
The value of

k

1, 2, 3, 4, 5 . . . . . . n

.
Putting k=1,

a_{1} = 2 \times 1 + 1

\Rightarrow a_{1} = 3

Putting k=2,

a_{2} = 2 \times 2 + 1

\Rightarrow a_{2} = 5

Similarly,

a_{3} = 7

a_{4} = 9

and so on.
The sum will be

3 + 5 + 7 + 9 + \dots

Common difference, d=5-3

\Rightarrow

d=2
Sum of n terms in a series is

S_{n}

\frac{n}{2} [2 a_{1} + (n - 1) \times d]

a_{1} = 3, d = 2

.
Substitute the values,

S_{n}

\frac{n}{2} [2 \times 3 + (n - 1) \times 2]

S_{n}

\frac{n}{2}

[6 + 2 n - 2]

S_{n}

\frac{n}{2} [4 + 2 n]

S_{n}

n (n + 2)

Hence, required sum is

n (n + 2)

.
Therefore, option 3 is correct.

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