The coefficient of x^5 in the expansion of (1 | vedclass

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(A) The coefficient of $x^r$ in the expansion of $(1+x)^n$ is ${ }^n C_r$. The coefficient of $x^5$ in the given expression is the sum of the coeffici

Vedclass · Accepted Answer

${ }^{31} C_6-{ }^{21} C_6$

Vedclass · Answer

(A) The coefficient of $x^r$ in the expansion of $(1+x)^n$ is ${ }^n C_r$. The coefficient of $x^5$ in the given expression is the sum of the coefficients of $x^5$ in each term: ${ }^{21} C_5 + { }^{22} C_5 + { }^{23} C_5 + \ldots + { }^{30} C_5$. Using the identity ${ }^n C_r + { }^n C_{r-1} = { }^{n+1} C_r$,we can rewrite this as: ${ }^{21} C_5 + { }^{22} C_5 + \ldots + { }^{30} C_5 = ({ }^{21} C_6 + { }^{21} C_5 + { }^{22} C_5 + \ldots + { }^{30} C_5) - { }^{21} C_6$. Applying the identity repeatedly: ${ }^{21} C_6 + { }^{21} C_5 = { }^{22} C_6$. ${ }^{22} C_6 + { }^{22} C_5 = { }^{23} C_6$. Continuing this process up to the last term: ${ }^{30} C_6 + { }^{30} C_5 = { }^{31} C_6$. Thus,the sum is ${ }^{31} C_6 - { }^{21} C_6$.

The coefficient of $x^5$ in the expansion of $(1+x)^{21}+(1+x)^{22}+\ldots+(1+x)^{30}$ is

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