The sum 1^2-2.3^2+3.5^2-4.7^2+5.9^2- +15.29^2 is ........ - Vidyadip

Question

The sum $1^2-2.3^2+3.5^2-4.7^2+5.9^2-\ldots +15.29^2$ is $.......$.

✓

Answer

d
Separating odd placed and even placed terms we get

$S =\left(1 \cdot 1^2+3 \cdot 5^2+\ldots .15 \cdot(29)^2\right)-\left(2 \cdot 3^2+4.7^2\right.+\ldots .+14 \cdot(27)^2$

$S =\sum \limits_{ n =1}^8(2 n -1)(4 n -3)^2-\sum_{ n =1}^7(2 n )(4 n -1)^2$

Applying summation formula we get

$=29856-22904=6952$

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

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The sum of the solutions of the equation $\left| {\sqrt x - 2} \right| + \sqrt x \left( {\sqrt x - 4} \right) + 2 = 0\left( {x > 0} \right)$ is equal to

Let the set $\mathrm{S}=\{2,4,8,16, \ldots . .512\}$ be partitioned into $3$ sets $A, B, C$ with equal number of elements such that $\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}=\mathrm{S}$ and $\mathrm{A} \cap \mathrm{B}=\mathrm{B} \cap \mathrm{C}=\mathrm{A} \cap \mathrm{C}=\phi$. The maximum number of such possible partitions of $S$ is equal to :

If ${T_n} = \,({n^2} + 1)n!\, \,{S_n} = \,{T_1} + {T_2} + {T_3} + ......{T_n}$ Let $\frac{{{T_{10}}}}{{{S_{10}}}} = \frac{a}{b}$ where $a$ & $b$ are realtively prime natural numbers, then the value of ($b - a$) is

A population $p(t) $ of $ 1000 $ bacteria introduced into nutrient medium grows according to the relation $p(t) = 1000 + {{1000t} \over {100 + {t^2}}}$. The maximum size of this bacterial population is

Four fair dice are thrown independently $27$ times. Then the expected number of times, at least two dice show up a three or a five, is

If $A$ is the area in the first quadrant enclosed by the curve $C: 2 x^2-y+1=0$, the tangent to $C$ at the point $(1,3)$ and the line $x+y=1$, then the value of $60 A$ is

$110$ triangles can be formed by joning $10$ points as vertices in which $n$ points are collinear. Then the value of $n$ is

Let $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$ . Let $\vec c$ be vector such that $\left| {\vec c - \vec a} \right| = 3,\;\left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right| = 3$ and the angle between $\vec c$ and $\vec a \times \vec b$ be $30^\circ $ . Then $\vec a \cdot \vec c$ is equal to :

Let $S =(0,2 \pi)-\left\{\frac{\pi}{2}, \frac{3 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}\right\}$. Let $y =$ $y ( x ), x \in S$, be the solution curve of the differential equation $\frac{ dy }{ dx }=\frac{1}{1+\sin 2 x }, y \left(\frac{\pi}{4}\right)=\frac{1}{2}$. if the sum of abscissas of all the points of intersection of the curve $y=y(x)$ with the curve $y=\sqrt{2} \sin x$ is $\frac{k \pi}{12}$, then $k$ is equal to

Let for $n =1,2, \ldots \ldots, 50, S _{ a }$ be the sum of the infinite geometric progression whose first term is $n ^{2}$ and whose common ratio is $\frac{1}{(n+1)^{2}}$. Then the value of $\frac{1}{26}+\sum\limits_{n=1}^{50}\left(S_{n}+\frac{2}{n+1}-n-1\right)$ is equal to