The coefficient of x 9 in the expansion of ∏ r = 1 100 1 + x r is equal to

Given ∏ r = 1 100 1 + x r = 1 + x 1 + x 2 . . . . . . . . . . . . 1 + x 100 So there will be 8 possible cases i.e. = x 9 , x 1 + 8 , x 2 + 7 , x 3 + 6 , x 4 + 5 , x 1 + 2 + 6 , x 1 + 3 + 5 , x 2 + 3 + 4 Now we can see that the coefficient of each will be 1 So coefficient of x 9 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8

E M B I B E

Gamma Question Bank for Engineering Mathematics>Chapter 6 - Binomial Theorem>Assignment>Q 46

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$2^{60}$ when divided by $7$ leave the remainder

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Important Questions on Binomial Theorem

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Gamma Question Bank for Engineering Mathematics>Chapter 6 - Binomial Theorem>Assignment>Q 47

Let

R = (5 \sqrt{5} + 11)^{2 n + 1}

and

f = R - [R]

where

[\cdot]

denotes the greatest integer function. The value of R is:

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Gamma Question Bank for Engineering Mathematics>Chapter 6 - Binomial Theorem>Assignment>Q 48

Consider the following statements
$S_{1}$ : The total number of terms in ${(x^{2} + 2 x + 1)}^{10}$ is $21$ .
$S_{2}$ : The coefficient of $x^{10}$ in ${(x^{2} + \frac{1}{x})}^{20}$ is ${}^{20}C_{10}$ .

$S_{3} :$ The middle term in the expansion of $(1 + x)^{12}$ is ${}^{12}C_{6} x^{6}$ .

$S_{4}$ : If the coefficients of the fifth and ninth term in the expansion of $(1 + x)^{n}$ are same then $n = 12$ .
Now identify the correct combination of true statements.