1+4/15+4 · 10/15 · 30+4 · 10 · 16/15 · 30 · 45+ ∞ = [AP EAPCET 23 May 2025 S2]

Correct Answer is : (53)23≤ft(5/3)²/3}(35)32

Correct Answer is : (53)23≤ft(5/3)²/3}(35)32

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Sequences and Series

Section: Mathematics

Question : 1 of 25

Marks: +1, -0

1+\frac{4}{15}+\frac{4 \cdot 10}{15 \cdot 30}+\frac{4 \cdot 10 \cdot 16}{15 \cdot 30 \cdot 45}+\ldots \infty =

Solution:

1+\frac{4}{15}+\frac{4 \cdot 10}{15 \cdot 30}+\frac{4 \cdot 10 \cdot 16}{15 \cdot 30 \cdot 45}+\ldots +\infty

We know that

(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2

+\frac{n(n-1)(n-2)}{3!}x^3+\ldots

On comparing with given expansion, we get

nx=\frac{4}{15}\Rightarrow x=\frac{4}{15n}

and

\frac{n(n-1)}{2!}x^2=\frac{4 \cdot 10}{15 \cdot 30}=\frac{40}{450}=\frac{4}{45}

Using Eq. (i),

\frac{n(n-1)}{2!}\left(\frac{4}{15n}\right)^2=\frac{4}{45}

\Rightarrow \frac{n(n-1)}{2}\times\frac{4\times4}{15n\times15n}

\Rightarrow \frac{n(n-1)}{2}\times\frac{16}{225n^2}

=\frac{4}{45}

n=\frac{-2}{3}

\text{ Also, } x=\frac{4}{15}\left(\frac{-3}{2}\right)=\frac{-2}{5}

\therefore (1+x)^n=\left(1-\frac{2}{5}\right)^{-\frac{2}{3}}

=\left(\frac{3}{5}\right)^{-\frac{2}{3}}=\left(\frac{5}{3}\right)^{\frac{2}{3}}

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