New chat PYQ Library Doubts Rank report

Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →

Binomial Theorem question

2025 · 3 Apr · Shift 1 · Q41

Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.

PYQ Library
JEE Main
Mathematics
Binomial Theorem
2025 · 3 Apr · Shift 1 · Q41

Binomial Theorem question

2025 · 3 Apr · Shift 1 · Q41

JEE MainMathematicsBinomial TheoremMCQ+4 / −1

If

\sum\limits_{r=1}^9\left(\frac{r+3}{2^r}\right) \cdot{ }^9 C_r=\alpha\left(\frac{3}{2}\right)^9-\beta, \alpha, \beta \in \mathbb{N}

, then

(\alpha+\beta)^2

is equal to

A
27
B
81
C
18
D
9

View written solutionFree

Correct answer: B

We need to evaluate $S=\sum_{r=1}^9 \left(\frac{r+3}{2^r}\right){^9C_r}.$ We are told that $S=\alpha\left(\frac{3}{2}\right)^9-\beta,$ where $\alpha,\beta\in\mathbb N$ .
Split the sum: $S=\sum_{r=1}^9 \frac{r}{2^r}{^9C_r}+3\sum_{r=1}^9 \frac{1}{2^r}{^9C_r}.$ So let

\qquad S_2=\sum_{r=1}^9 {^9C_r}\left(\frac12\right)^r.$$ Then $$S=S_1+3S_2.$$ 3. Evaluate $S_2$ using the binomial theorem: $$\sum_{r=0}^9 {^9C_r}x^r=(1+x)^9.$$ Putting $x=\frac12$, $$\sum_{r=0}^9 {^9C_r}\left(\frac12\right)^r=\left(1+\frac12\right)^9=\left(\frac32\right)^9.$$ Hence, $$S_2=\left(\frac32\right)^9-1$$ since the $r=0$ term is $1$. 4. Evaluate $S_1$ using the standard identity $$\sum_{r=0}^n r{^nC_r}x^r=nx(1+x)^{n-1}.$$ For $n=9$ and $x=\frac12$, $$\sum_{r=0}^9 r{^9C_r}\left(\frac12\right)^r=9\cdot \frac12\left(1+\frac12\right)^8.

Thus,

Now simplify: $\frac{9}{2}\left(\frac32\right)^8=3\left(\frac32\right)^9.$ So, $S_1=3\left(\frac32\right)^9.$

Therefore, $S=S_1+3S_2=3\left(\frac32\right)^9+3\left[\left(\frac32\right)^9-1\right].$ So, $S=6\left(\frac32\right)^9-3.$ Comparing with $S=\alpha\left(\frac32\right)^9-\beta,$ we get $\alpha=6,\qquad \beta=3.$
Hence, $(\alpha+\beta)^2=(6+3)^2=9^2=81.$
Matching with the options, the correct answer is: $\boxed{81}$ which is option B.

More from Binomial Theorem

Let $(1 + x + x^{2})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{20} x^{20}$ . If $(a_{1} + a_{3} + a_{5} + \dots + a_{19}) - 11 a_{2} = 121 k$ , then $k$ is equal to .2025 · Numerical
For an integer $n \geq 2$ , if the arithmetic mean of all coefficients in the binomial expansion of $(x + y)^{2 n - 3}$ is 16 , then the distance of the point $P (2 n - 1, n^{2} - 4 n)$ from the line $x + y = 8$ is2025 · MCQ
In the expansion of $(32 + \frac{1}{3 3})^{n}, n \in N$ , if the ratio of $1 5^{th}$ term from the beginning to the $1 5^{th}$ term from the end is $\frac{1}{6}$ , then the value of $^{n} C_{3}$ …2025 · MCQ
If $1^{2} \cdot (^{15} C_{1}) + 2^{2} \cdot (^{15} C_{2}) + 3^{2} \cdot (^{15} C_{3}) + \dots + 1 5^{2} \cdot (^{15} C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}$ , where $m, n, k \in N$ , then $m + n + k$ …2025 · MCQ
The remainder when $((64)^{(64)})^{(64)}$ is divided by 7 is equal to2025 · MCQ
The sum of the series $2 \times 1 \times^{20} C_{4} - 3 \times 2 \times^{20} C_{5} + 4 \times 3 \times^{20} C_{6} - 5 \times 4 \times^{20} C_{7} + \dots\dots + 18 \times 17 \times^{20} C_{20}$ ,…2025 · Numerical
The number of integral terms in the expansion of $(5^{\frac{1}{2}} + 7^{\frac{1}{8}})^{1016}$ is:2025 · MCQ
The product of the last two digits of $(1919)^{1919}$ is2025 · Numerical