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  • <img alt="\sum_{r=0}^{300} a_r x^r=\left(1+x+x^2+x^3\right)^{100}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Csum_%7Br%3D0%7D%5E%7B300%7D%20a_r%20x%5Er%3D%5Cleft%281&plus;x&plu

\sum_{r=0}^{300} a_r x^r=\left(1+x+x^2+x^3\right)^{100}  If a =\sum_{r=0}^{300} a_r \text { then } \sum_{r=0}^{300} r \cdot a_r is equal to

 

Option: 1

300a


Option: 2

100a


Option: 3

150a


Option: 4

75a


Answers (1)

best_answer

\sum_{r=0}^{300} a_r x^r=\left(1+x+x^2+x^3\right)^{100}

Clearly, ‘ar’ is the coefficient of xr in the expansion of (1 + x + x2 + x3)100

Replacing x →1/x in the given equation, we get

\begin{aligned} & \sum_{r=0}^{300} a_r\left(\frac{1}{x}\right)^r=\frac{1}{x^{300}}\left(x^3+x^2+x+1\right)^{100} \\ & \sum_{r=0}^{300} a_r x^{300-r}=\left(1+x+x^2+x^3\right)^{100} \end{aligned}

Here ar represents of coefficient. of x^{300-r} \text { in }\left(1+x+x^2+x^3\right)^{100}

Thus a_r=a_{300-r}

\begin{aligned} & \mathrm{I}=\sum_{r=0}^{300} r \cdot a_r=\sum_{r=0}^{300}(300-r) a_{300-r} \\ & =\sum_{r=0}^{300}(300-r) a_r=\sum_{r=0}^{300} a_r-\sum_{r=0}^{300} r \cdot a_r \\ & 2 I=300 a \\ & I=150 a \end{aligned}

 

Posted by

Irshad Anwar

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