First slide

Binomial theorem for positive integral Index

Question

In the expansion of ${(1 + x + \frac{7}{x})}^{11}$ , the term not containing x is

Moderate

A

$\sum_{r = 0}^{5}^{11} {C_{r}}^{11 - r} {C_{2 r} 7}^{r}$
B

$\sum_{r = 1}^{5}^{11} {C_{r}}^{11 - r} C_{11 - 2 r} 7^{r}$
C

$\sum_{r = 0}^{5}^{11} {C_{r}}^{11 - r} C_{11 - 2 r} 7^{r}$
D

$\sum_{r = 1}^{5} {}^{11}{{}^{c}r}$

Solution

${(1 + x + \frac{7}{x})}^{11} = {(\frac{7 + x + x^{2}}{x})}^{11}$
So, we have to find the coefficient of x¹¹ in ${(7 + x + x^{2})}^{11}$
Now $\begin{array}{l} {(7 + x + x^{2})}^{11} \end{array}$
$\begin{array}{l} {= [(7 + x) + x^{2})]}^{11} \\ =^{11} C_{0} (7 + x)^{11} +^{11} C_{1} (7 + x)^{10} x^{2} +^{11} C_{2} (7 + x)^{9} x^{4} \dots + \dots \end{array}$
$\begin{array}{rcl} ∴ Required coefficient =^{11} C_{0} +^{i 1} C_{1} \times^{10} C_{9} \times 7 +^{11} C_{2} \times^{9} C_{7} \times 7^{2} +^{11} C_{3} \times^{8} C_{5} \times 7^{3} +^{11} C_{4} \times^{7} C_{3} \times 7^{4} +^{11} C_{5} \times^{6} C_{1} \times 7^{5} \\ = & \sum_{r = 0}^{5}^{11} C_{r}^{11 - r} C_{11 - 2 r} 7^{r} \end{array}$

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