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  • Find the cube root of 318Option: 1 6.71Option: 2 6.79Option

Find the cube root of 318

Option: 1 6.71

Option: 2 6.79

Option: 3 6.83

Option: 4 6.88

Answers (1)

best_answer

 

 

Binomial Theorem for any Index -

Binomial Theorem for any  Index

\\\text{The given series is}\\\mathrm{(1+x)^n=^nC_0+^nC_1x+^nC_2x^2+....+^nC_rx^r+....+^nC_nx^n}\\\text{for negative or fractional Index}\\\\\mathrm{(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+......+\frac{n(n-1)(n-2).....(n-r+1)}{r!}x^r....\infty}

Note:

  1. If n is negative or fractional index then this condition |x| < 1 is essential.

  2. There is an infinite number of terms in the expansion of (1 + x)n, when n is negative or fractional index.

 

If the first term is not unity and the index of the binomial is either a negative integer or a fraction, then we expand as follows:

 \begin{aligned}(x+a)^{n} &=\left\{a\left(1+\frac{x}{a}\right)\right\}^{n}+a^{n}\left(1+\frac{x}{a}\right)^{n} \\ &=a^{n}\left\{1+n \frac{x}{a}+\frac{n(n-1)}{2 !}\left(\frac{x}{a}\right)^{2}+\cdots\right\} \\ &=a^{n}+n a^{n-1} x+\frac{n(n-1)}{2 !} a^{n-2} x^{2}+\cdots \end{aligned}

 

The above expansion is valid when \left | \frac{x}{a} \right |<1.

 

-

 

\\\begin{array}{c}{(318)^{1 / 3}=\left(7^{3}-25\right)^{1 / 3}=7\left(1-\frac{25}{7^{3}}\right)^{1 / 3}} \\\\ {=7\left(1- \frac{25}{3 \times 343}+\frac{1 \times 2}{3 \times 3 \times 2!}\left(\frac{25}{343}\right)^{2}+\ldots\right)=7(1-0.0243)=6.83}\end{array}

Hence, Option C is correct

Posted by

Suraj Bhandari

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