Summation by Sigma Method

Q.

The sum of the product of the integers 1, 2, 3, ...., n taken two at a time is

1. $\frac{n(n+1)}{24}(3n^2-n-2)$

2. $\frac{n^2(n+1{)}^2}{2}(n^2+1)$

3. $\frac{n(n+1)}{18}(n^2-n-2)$

4. $\frac{n(n+1)(n+2)(n+3)}{24}$

Q. If $S_n={\sum }_1^{4n}(-1{)}^{\frac{k(k+1)}{2}}{k}^2$. Then, $S_n$ can take value(s)

1. 1056
2. 1088
3. 1120
4. 1332

Q. The sum of first 9 terms of the series $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots \dots$ is

1. 71
2. 96
3. 142
4. 192

Q. IF $S_n={\sum }_{r=0}^n\frac{1}{{}^n{C}_r}$ and $t_n={\sum }_{r=0}^n\frac{r}{{}^n{C}_r},$ then $\frac{t_n}{S_n}$ is equal to

1. $n-1$
2. $\frac{2n-1}{2}$
3. $\frac{1}{2}n-1$
4. $\frac{1}{2}n$

Q.

A series in G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying odd places, then the common ratio will be equal to

1. 2

2. 5

3. 3

4. 4

Q.

Find the sum of the first 15 terms of the series 3 + 5 + 7 + 9 +. . . . . . n terms.

___

Q.

Find the sum of the first n natural numbers.

1. 

2. 

3. 

4. 

Q.

If the sum of the n terms of G.P. is S product is P and sum of their inverse is R, than $P^2$ is equal to

1. 

2. 

3. 

4. 

Q. If $2+5+8+11+\cdots \text{upto}n\text{terms}=610$, then the value of $n$ is

1. $19$
2. $22$
3. $21$
4. $20$

Q.

The sum of the product of the integers 1, 2, 3, ...., n taken two at a time is

1. $\frac{n(n+1)}{24}(3n^2-n-2)$

2. $\frac{n^2(n+1{)}^2}{2}(n^2+1)$

3. $\frac{n(n+1)}{18}(n^2-n-2)$

4. $\frac{n(n+1)(n+2)(n+3)}{24}$