Question

find the 11th term from the beginning and 11th terms end in the expansion of (2x-1/x^2)^25​

plz ans its urgent

Answer 1

The 11th term from the beginning and end are  $-25C_{15} \frac{2^{10} }{x^{20} }$ and $25C_{10}\frac{2^{15} }{x^{5} }$ respectively.

• A series in mathematics is essentially a description of the process of successively adding an infinite number of quantities to a specified initial quantity. A significant component of calculus and its generalisation, mathematical analysis, is the study of series.
• Most branches of mathematics use series, including combinatorics, where generating functions are used to explore finite structures. In addition to being widely utilised in mathematics, infinite series are also used extensively in physics, computer science, statistics, and finance, among other quantitative fields.

Here, according to the given information, we are given that,

$(2x-\frac{1}{x^{2} } )^{25}$

Now, since the given series is raised to the power 25, these series must contain 26 terms.

Now, we need to find the eleventh term from the beginning and the eleventh term from the end of this series.

Now, we see that, the eleventh term from the beginning is $(26-11+1) = 16$ th term when we see it from the end of the series that has been given.

Then, we get,

The 16th term = $25C_{15}2x^{10}( -\frac{1}{x^{2} }) ^{15}$

= $-25C_{15} \frac{2^{10} }{x^{20} }$

Similarly, the eleventh term is,

$25C_{10}2x^{10}( -\frac{1}{x^{2} }) ^{10}\\=25C_{10}\frac{2^{15} }{x^{5} }$

Hence, the 11th term from the beginning and end are  $-25C_{15} \frac{2^{10} }{x^{20} }$ and $25C_{10}\frac{2^{15} }{x^{5} }$ respectively.

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