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**Hint:**The given sum 1+3+5+… is the sum of a sequence, which is in AP since the difference between consecutive terms is the same. Now we know the sum of n terms of AP is $\dfrac{n}{2}\left[ 2a+(n-1)d \right]$ where a is the first term, d is a common difference, and n is the number of terms.

Hence with the help of this formula and the given condition we can find required n.

**Complete step-by-step solution:**Now the given series is $1+3+5+……….$

Here we know the first term is 1.

Difference between any two consecutive terms is \[5-3=3-1=2\]

Now let us try to find the sum of n terms of this series. We know sum of AP is given by $\dfrac{n}{2}\left[ 2a+(n-1)d \right]$

Here we have $a=1$ and $d=2$. Hence sum of n terms will be

$\dfrac{n}{2}\left[ 2(1)+(n-1)(2) \right]$

$=\dfrac{n}{2}\left[ 2+2n-2 \right]$

\[=\dfrac{n}{2}\times \left[ 2n \right]\]

\[={{n}^{2}}\]

Hence now we have the sum $1+3+5+…n$ is \[={{n}^{2}}\]

Now we have to find n such that the sum of the given series is greater than in 1357.

Now let us check the options.

If $n = 37$ then ${{n}^{2}}=37\times 37=1369$

If $n = 36$ then ${{n}^{2}}=36\times 36=1296$

Now for n = 36 we get ${{n}^{2}}=1296<1357$ and for n = 37 we get ${{n}^{2}}=1369$

Hence, the minimum number such that the sum of $1+3+5+…n$ is exceeding 1357 is 1369 and this will happen for $n = 37.$

Hence, the number of terms required is 37.

Option b is the correct option.

**Note:**Since the sum of odd natural numbers is equal to \[{{n}^{2}}\]we just have to find the square root of sum for thee required n. If the options are not available by hook or crook we can substitute the value of n and find ${{n}^{2}}$ such that it satisfies the required condition. In this case, the condition is to find least n such that${{n}^{2}} >1357$. Hence all we have to do is find n such that ${{n}^{2}}> 1357$ and ${{(n-1)}^{2}}< 1357$.

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