Question

1+ 22 + 24 + 26 + … + 248 is equal to:
(a) 2⁴⁹−1
(b) (2⁴⁹−1)/3
(c) 2⁵⁰−1
(d) (2⁵⁰−1)/3
(e) None of the Above (Specify Correct Answer)

Answer 1

Answer:

The infinite series whose terms are the natural numbers 1 + 2 + 3 + 4 + ⋯ is a divergent series. The nth partial sum of the series is the triangular number

{\displaystyle \sum _{k=1}^{n}k={\frac {n(n+1)}{2}},}{\displaystyle \sum _{k=1}^{n}k={\frac {n(n+1)}{2}},}

Step-by-step explanation:

Answer 2

Answer:

(e) None of the Above (Correct Answer is 15391)

Step-by-step explanation:

Given, series is 1 + 22 + 24 + 26 + … + 248

To Find: The number equivalent to above mentioned series.

Explanation:

1 + 22 + 24 + 26 + … + 248

= 1 + 2(11 + 12 + 13 + ... + 124)

• Here, 11, 12, 13, 14,...,124 is an Arithmetic Progression.

• Now comes the question, “What is an Arithmetic Progression?”

• The answer comes as: An Arithmetic Progression (A.P.) is an arithmetic sequence (of numbers) in which the difference between two consecutive terms is always constant.

        For Example, 2,4,6,8,10 are in Arithmetic Progression.

• The First term of an A.P. is typically denoted by “a”.

• While the Last term of an A.P. is denoted by “l”.

• The common difference of an A.P. is the constant difference between two consecutive terms (as mentioned earlier in the definition of arithmetic progression) and is typically denoted by “d”.

• Here, we will have to calculate the number of terms present in our concerned A.P. and then, using that information, we will have to calculate the sum of all terms.

1. To find a certain nth term of an A.P., we need to use the formula of $Tn=[a+(n-1)d]$.
2. Here, we know the first and last terms of our A.P. and also, we know the common difference. We will put the known values in the above mentioned formula, and treat it as an equation. Then, we will solve the equation for "x" and we will be obtain the total number of terms in the A.P.
3. To find the sum of n terms of an A.P., we need to use the formula of    $Sn=\frac{n}{2}[2a+(n-1)d]$

• Here, (a = 11), (l = 124) and (d = 1)

Therefore, using Formula (1), we get,

$124=[11+(n-1)1]\\or, 124=11+(n-1)\\or, 124-11=(n-1)\\or, (n-1)=113\\or,n=(113+1)=114$

Now using (n = 114), in Formula (3), we get,

$Sn=\frac{114}{2}[(2*11)+(114-1)1]\\ or, Sn=57(22+113)\\or, Sn=57*135\\or, Sn=7695$

Therefore, 1 + 22 + 24 + 26 + … + 248

= 1 + 2(11 + 12 + 13 + ... + 124)

= 1 + 2(Sn)

= 1 + (2 * 7695)

= 1 + 15390

=15391

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