Download App

Some Special Series — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsSome Special Series5 Marks
Question
Find the sum of the series whose $n^{th}$ term is:
$2n^3 + 3n^2 - 1$
✓

Answer

$T_n = 2n^3 + 3n^2 - 1$ Let $S_n $ be the sum of n terms of the given series.
Now,
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}(2\text{k}^2+3\text{k}-1)$
$\Rightarrow\text{S}_\text{n}=2\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2+3\sum\limits^{\text{n}}_{\text{k}=1}\text{k}-\sum\limits^{\text{n}}_{\text{k}=1}1$
$\Rightarrow\text{S}_\text{n}=\frac{2\text{n}^2(\text{n}+1)^2}{4}+\frac{3\text{n}(\text{n}+1)(2\text{n}+1)}{6}-\text{n}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^2(\text{n}+1)^2}{2}+\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{2}-\text{n}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^2(\text{n}+1)^2+\text{n}(\text{n}+1)(2\text{n}+1)-2\text{n}}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^2(\text{n}^2+1+2\text{n})+(\text{n}^2+\text{n})(2\text{n}+1)-2\text{n}}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^2(\text{n}^2+1+2\text{n})+(2\text{n}^3+\text{n}^2+2\text{n}^2+\text{n})-2\text{n}}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^4+4\text{n}^2+4\text{n}^3-\text{n}}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}^3+4\text{n}+4\text{n}^2-1)}{2}$

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.(5 Marks)" from Some Special Series→Some Special Series — all question types→Maths STD 11 Science question papers→Maharashtra Board STD 11 Science question papers→Maharashtra Board question paper generator→

Similar questions

Evaluate the following limits:
$\lim _{x \rightarrow \pi}\left[\frac{\sqrt{5+\cos x-2}}{(\pi-x)^2}\right]$
→
Find the equations to the straight lines which pass through the point $(h, k)$ and are inclined at angle $\tan^{-1}$ m to the straight line $y = mx + c.$
→
Prove that $\frac{1}{2}\tan\Big(\frac{\text{x}}{2}\Big)+\frac{1}{4}\tan\Big(\frac{\text{x}}{4}\Big)+...+\frac{1}{2^{\text{n}}}\tan\Big(\frac{\text{x}}{2^{\text{n}}}\Big)=\frac{1}{2^{\text{n}}}\cot\Big(\frac{\text{x}}{2^{\text{n}}}\Big)-\cot\text{x}$ for all $\text{n}\in\text{N}$ and $0<\text{x}<\frac{\pi}{2}$
→
If the line $2x - y + 1 = 0$ touches the circle at the point $(2, 5)$ and the centre of the circle lies on the line $x + y - 9 = 0.$ Find the equation of the circle.
→
Differentiate the following from first principle$\tan\text{x}^2$
→
$5^{2n} - 1$ is divisible by $24$ for all $\text{n}\in\text{N}.$
→
Find the eccentricity, coordinates of the foci, equation of the directrices and lenght of the latus-rectum of the hyperbola
$4\text{x}^{2}-3\text{y}^{2}=36$
→
Show that the product of the lengths of the perpendicular segments drawn from the foci

to any tangent line to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ is equal to 16 .

→
Prove that the following sets of three lines are concurrent:
15x - 18y + 1 = 0, 12x + 10y - 3 = 0 and 6x + 66y - 11 = 0
→
Find the coording of the centre of the circle inscribed in a triangle whose vertics are (-36, 7), (20, 7) and (0, -8).
→