Answer

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Hint: We have to find the cube root of 64 using the given series. Observe the series given by adding the successive terms and form a new series. So the cube root is found by subtracting 64 with successive numbers from the series till the answer is zero.

Complete step-by-step answer:

We have series given as

1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397,.................(1)

Let us first observe the relation in the given series.

First term = 1

Second term + first term = 7+1 = 8

Third term + second term + first term = 27

Similarly,

Fourth + third + second + first = 64.

Hence, if we add the successive terms and write a new series as

${{a}_{1}},{{a}_{1}}+{{a}_{2}},{{a}_{1}}+{{a}_{2}}+{{a}_{3}},{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}},.........$

Where ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}}.......$ are the terms of the series given in equation (1),

Where ${{a}_{1}}=1,{{a}_{2}}=7,{{a}_{3}}=19,{{a}_{4}}=37\text{ and so on}\text{.}$

Hence, new series we get;

1, 8, 27, 64, 125, 216.........(2)

Now, observing the above series, we observe that all terms are the cubes of the natural numbers i.e. ${{\left( 1 \right)}^{3}},{{\left( 2 \right)}^{3}},{{\left( 3 \right)}^{3}},{{\left( 4 \right)}^{3}},{{\left( 5 \right)}^{3}}...........$

Now, as we have to determine cube root of 64 by the successive difference of the terms of the given series in equation (1).

Now, think opposite of it that 64 is written by adding the first four times of the first series as $1+7+19+37.$

Therefore, we can subtract the terms of series (1) from 64 from starting and at the point where (64) given number becomes zero; we get cube root by number of times successive difference taken place in following way;

Step 1: 64-1 = 63

Step 2: 63-7 = 56

Step 3: 56-19 = 37

Step 4: 37-37 = 0

Number of successive differences that take place to get zero is 4.

Hence, the cube root of 64 is 4.

Note: Students might not use the given series and write the cube root of 64 using any other known method. But the question here demands a particular method to find out the cube root.

Complete step-by-step answer:

We have series given as

1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397,.................(1)

Let us first observe the relation in the given series.

First term = 1

Second term + first term = 7+1 = 8

Third term + second term + first term = 27

Similarly,

Fourth + third + second + first = 64.

Hence, if we add the successive terms and write a new series as

${{a}_{1}},{{a}_{1}}+{{a}_{2}},{{a}_{1}}+{{a}_{2}}+{{a}_{3}},{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}},.........$

Where ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}}.......$ are the terms of the series given in equation (1),

Where ${{a}_{1}}=1,{{a}_{2}}=7,{{a}_{3}}=19,{{a}_{4}}=37\text{ and so on}\text{.}$

Hence, new series we get;

1, 8, 27, 64, 125, 216.........(2)

Now, observing the above series, we observe that all terms are the cubes of the natural numbers i.e. ${{\left( 1 \right)}^{3}},{{\left( 2 \right)}^{3}},{{\left( 3 \right)}^{3}},{{\left( 4 \right)}^{3}},{{\left( 5 \right)}^{3}}...........$

Now, as we have to determine cube root of 64 by the successive difference of the terms of the given series in equation (1).

Now, think opposite of it that 64 is written by adding the first four times of the first series as $1+7+19+37.$

Therefore, we can subtract the terms of series (1) from 64 from starting and at the point where (64) given number becomes zero; we get cube root by number of times successive difference taken place in following way;

Step 1: 64-1 = 63

Step 2: 63-7 = 56

Step 3: 56-19 = 37

Step 4: 37-37 = 0

Number of successive differences that take place to get zero is 4.

Hence, the cube root of 64 is 4.

Note: Students might not use the given series and write the cube root of 64 using any other known method. But the question here demands a particular method to find out the cube root.

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