The value of \({{\rm{\Delta }}^{10}}\left[ {\left( {1 - ax} \right)\left( {1 - b{x^2}} \right)\left( {1 - c{x^3}} \right)\left( {1 - d{x^4}} \right)} \right]\) is

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  1. abcd (10!)
  2. abcd (9!)
  3. abcd (8!)
  4. abcd (7!)

Answer (Detailed Solution Below)

Option 1 : abcd (10!)
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Detailed Solution

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Concept:

Forward difference:

  • Suppose that a function y = f(x) is tabulated for the equally spaced arguments x0, x0 + h, x0 + 2h, …, x0 + nh giving the functional values y0, y1, y2, …, yn.
  • The constant difference between two consecutive values of x is called the interval of differences and is denoted by h.

 

The operator Δ defined by

Δy0 = y1 – y0

Δy0 = y2 – y1

………

………

Δyn – 1 = yn – yn – 1

Δ is called forward difference operator.

The first forward difference is Δyn = yn+1 – yn

The second forward difference is Δ2yi = Δy1 + i – Δyi

In general, nth forward difference of f is defined by

Δnyi = Δn – 1 yi + 1 – Δn – 1yi

For a continuous function, forward difference operator ‘Δ’ and differentiation operator \(\frac{d}{{dx}}\) are similar.

Calculation:

\({{\rm{\Delta }}^{10}}\left[ {\left( {1 - ax} \right)\left( {1 - b{x^2}} \right)\left( {1 - c{x^3}} \right)\left( {1 - d{x^4}} \right)} \right]\)

The order of given expression is 10 and we need to find 10th forward difference.

Similar to the differentiation, 10th forward difference of the terms lower than 10th order are zero.

So, the given expression is equivalent to

\({{\rm{\Delta }}^{10}}\left[ {abcd{x^{10}} + \ldots } \right]\)

= abcd (10!)

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