Download App

Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits4 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+\ \dots+\text{n}^3}{(\text{n}-1)^4}$
✓

Answer

$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+3^3\ \dots+\text{n}^3}{(\text{n}-1)^4}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big[\frac12(\text{n})(\text{n}+1)\Big]^2}{(\text{n}-1)^4}$ $\bigg[1^3+2^3+3^3+\ \cdots+\text{n}^3=\Big(\frac12\text{n}(\text{n}+1)\Big)^2\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\Bigg(\frac{\frac14\text{n}^2\big(\text{n}^2+1+2\text{n}\big)}{(\text{n}-1)^4}\Bigg)$
$=\frac14\lim\limits_{\text{n}\rightarrow\infty}\bigg(\frac{\text{n}^4+\text{n}^2+2\text{n}^3}{(\text{n}-1)^2(\text{n}-1)^2}\bigg)$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$
$=\frac{1}{4}\lim\limits_{\text{n}\rightarrow{\infty}}\Big(\frac{\text{n}^4+\text{n}^2+2\text{n}^3}{\text{n}^4+\text{n}^2-2\text{n}^3+\text{n}^2+1-2\text{n}-2\text{n}^3-2\text{n}+4\text{n}^2}\Big)$
$=\frac{1}{4}\lim\limits_{\text{n}\rightarrow{\infty}}\frac{\Big(1+\frac{1}{​​\text{n}^2}+\frac{2}{​​\text{n}}\Big)}{\Big(1+\frac{1}{\text{n}^2}+\frac{2}{\text{n}}+\frac{1}{\text{n}^2}+\frac{1}{\text{n}^4}-\frac{2}{\text{n}^3}-\frac{2}{\text{n}}-\frac{2}{\text{n}^3}+\frac{4}{\text{n}^2}\Big)}$
$=\frac{1}{4}\Big(\frac14\Big)$
$=\frac14$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "Solve the Following Question.(4 Marks)" from Limits→Limits — all question types→Maths STD 11 Science question papers→Maharashtra Board STD 11 Science question papers→Maharashtra Board question paper generator→

Similar questions

Prove by method of induction $15^{2 n-1}+1$ is divisible by 16 , for all $n \in N$.
→
Solve the following system of equations in R.
$|\text{x}+1|+|\text{x}|>3$
→
Sketch the graphs of the following trigonometric functions:
$\text{h(x)}=\cos^22\text{x}$
→
Evaluate the following limits: $\lim _{x \rightarrow a}\left[\frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}\right]$
→
If $\text{a}\cos2\text{x}+\text{b}\sin2\text{x}=\text{c}$ has $\alpha$ and $\beta$ as its roots, then prove that,
$\tan\alpha+\tan\beta=\frac{2\text{b}}{\text{a+c}}$
→
If a, b, c, d are in G.P., prove that:
$\big(\text{a}^2+\text{b}^2+\text{c}^2\big),\big(\text{ab}+\text{bc}+\text{cd}\big),\big(\text{b}^2+\text{c}^2+\text{d}^2\big)\text{ are in G.P.}$
→
Differentiate the following functions by the product rule and the other method and verify that the answer from both the methods is the same.$(3\sec\text{x}-4\text{cosec}\text{x})(-2\sin\text{x}+5\cos\text{x})$
→
Verify A(BC) = (AB)C in each of the following cases:

$A=\left[\begin{array}{cc}4 & -2 \\ 2 & 3\end{array}\right], B=\left[\begin{array}{cc}-1 & 1 \\ 3 & -2\end{array}\right]$ and $C=\left[\begin{array}{cc}4 & 1 \\ 2 & -1\end{array}\right]$

→
Find the equation of the circle the end points of whose diameter are the centres of the circles $x^2 + y^2 + 6x - 14y - 1 = 0$ and $x^2 + y^2 - 4x + 10y - 2 = 0$.
→
Let $f(x) = x^2 $ and $g(x) = 2x + 1$ be two real functions. Find $(f + g)(x), (f - g)(x), (fg)(x)$ and $\Big(\frac{\text{f}}{\text{g}}\Big)\text{x}$
→