Do you know its short-cut ?

$\sqrt[3]{1+3+5+.....+53}$

The difference between two consecutive terms is same therefore it an A.P. with common difference $2$ .

Let $53$ be the $nth$ term of the sequence.

$53=1+(n-1)2$

$n=27$

The sum of the sequence is $\frac{27}{2} (1+53)$ or ${27}^{2}$ .

We are required to find $\sqrt[3]{{27}^{2}}$ which is,

$\boxed{9}$

Note first that, in general for any integer $n \ge 1,$

$\displaystyle\sum_{k=1}^{n} (2k - 1) = 2\sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 = 2*\dfrac{n(n + 1)}{2} - n = n(n + 1) - n = n^{2}.$

In this case, since $53 = 2*27 - 1,$ we see that the given expression is

$(27^{2})^{\frac{1}{3}} = (27^{\frac{1}{3}})^{2} = 3^{2} = \boxed{9}.$

Sum of n odd numbers is $n^2$ . $\sqrt[3]{1+3+5+ \dots + 53}$ $\sqrt[3]{27^2}$ $\boxed{9}$

n= (53+1)/2 = 27 Sn = n/2(a +L) =27/2(1+53) =27(27) =9 ^3 according to the question the whole summation is under the power 1/3 so that 9 ^(3 *1/3) = 9

When numbers in arithmetic progression are added it can also be represented as the number of terms multiplied by the average of the smallest and largest number For ex:1+4+7=4 x 3 So this expression can be represented as 27 x 27

Here is the simple common sense solution:

For the sequence, 1+3+5+...+53.

• Number of terms of 1 = 1
• Number of terms of 3,5,...,53 = Number of terms of 2,4,...,52 = 52/2 = 26

• Total Number of Terms = 1 + 26 = 27

• Average Term Value = (1+53)/2 = (3+51)/2 = (5+49)/2 = 27

Sum of the sequence = the number of terms(27) multiplied by the average term value(27) = 27*27 = 729

$\sqrt[3]{729}$ = 9

lod dis is ez 53 = 27 th odd number 27^2 = 1+ 3 + 5.... + 53

27 = 3^3 (3^3)^2 = 3^6

(3^6 )^1/3 = 3 ^2

9