JEE Mains :

1. Sum of n terms of an A.P.

$$S_n=\frac{n}{2}[2a_1+(n-1)d]$$

where (a_1) is the first term, (d) is the common difference, and (n) is the number of terms.

2. Sum of n terms of a G.P.

$$S_n=\frac{a_1(r^n-1)}{r-1}$$

where (a_1) is the first term, (r) is the common ratio, and (n) is the number of terms.

3. Sum of n terms of an arithmetic-geometric series.

$$S_n=\frac{a_1(r^n-1)}{r-1}-\frac{d(n-1)}{2}$$

where (a_1) is the first term, (r) is the common ratio, (d) is the common difference, and (n) is the number of terms.

4. Sum of n terms of a harmonic series.

$$H_n=\sum _{i=1}^n\frac{1}{i}\approx \ln n+\gamma$$

where (\gamma \approx 0.57721) is the Euler-Mascheroni constant.

5. Sum of n terms of a telescoping series.

Telescoping series are series of the form

$$\sum _{i=1}^n(a_i-a_{i+1})$$

where (a_i) and (a_{i+1}) are consecutive terms of the series. The sum of a telescoping series can be found by simply subtracting the last term from the first term:

$$S_n=a_1-a_{n+1}$$

6. Sum of n terms of a binomial series.

The binomial series for ((1+x)^n) is given by

$$(1+x{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^k$$

where (\binom{n}{k}) is the binomial coefficient. The sum of the first n terms of the binomial series is known as the binomial sum and can be calculated using the formula

$$S_n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^k=(1+x{)}^n$$

7. Product of n terms of an A.P.

$$P_n=a_1\cdot a_2\cdot \dots =\frac{a_1[a_1+(n-1)d{]}^n}{a_1^n}$$

where (a_1) is the first term, (d) is the common difference, and (n) is the number of terms.

8. Product of n terms of a G.P.

$$P_n-a_1\cdot a_2\cdot a_3\cdot \dots \cdot a_n=a_1^n\cdot r^{n(n-1)/2}$$

where (a_1) is the first term, (r) is the common ratio, and (n) is the number of terms.

9. Product of n terms of an arithmetic-geometric series.

$$P_n=\frac{a_1(r^n-1)}{r-1}\cdot r^{\frac{n(n-1)}{2}}$$

where (a_1) is the first term, (r) is the common ratio, and (n) is the number of terms.

10. Product of n terms of a harmonic series.

There is no simple formula for the product of n terms of a harmonic series.

11. Find the sum of n terms of the sequence {a_n} defined by a_n = 3n - 1.

$$S_n=\sum _{i=1}^n(3i-1)=3\cdot \frac{n(n+1)}{2}-n$$

12. Find the sum of n terms of the sequence {a_n} defined by a_n = 1/n.

$$S_n=\sum _{i=1}^n\frac{1}{i}\approx \ln n+\gamma$$

13. Find the sum of n terms of the sequence {a_n} defined by a_n = n^2 - 1.

$$S_n=\sum _{i=1}^n(i^2-1)=\frac{n(n+1)(2n+2)}{6}$$

14. Find the sum of n terms of the sequence {a_n} defined by a_n = 2^n - 1.

$$S_n=\sum _{i=1}^n(2^n-1)=2^n-1$$

15. Find the sum of n terms of the sequence {a_n} defined by a_n = n!.

$$S_n=\sum _{i=1}^{n}n!=(n+1)!-1$$

CBSE Board Exams

1. Sum of n terms of an A.P.

$$S_n=\frac{n}{2}[2a_1+(n-1)d]$$

where (a_1) is the first term, (d) is the common difference, and (n) is the number of terms.

2. Sum of n terms of a G.P.

$$S_n=\frac{a_1(r^n-1)}{r-1}$$

where (a_1) is the first term, (r) is the common ratio, and (n) is the number of terms.

3. Sum of n terms of an arithmetic-geometric series.

$$S_n=\frac{a_1(r^n-1)}{r-1}-\frac{d(n-1)}{2}$$

where (a_1) is the first term, (r) is the common ratio, (d) is the common difference, and (n) is the number of terms.

4. Sum of n terms of a harmonic series.

$$H_n=\sum _{i=1}^n\frac{1}{i}\approx \ln n+\gamma$$

where (\gamma \approx 0.57721) is the Euler-Mascheroni constant.

5. Sum of n terms of a telescoping series.

Telescoping series are series of the form

$$\sum _{i=1}^n(a_i-a_{i+1})$$

where (a_i) and (a_{i+1}) are consecutive terms of the series. The sum of a telescoping series can be found by simply subtracting the last term from the first term:

$$S_n=a_1-a_{n+1}$$

6. Find the sum of n terms of the sequence {a_n} defined by a_n = 3n - 1.

$$S_n=\sum _{i=1}^n(3i-1)=3\cdot \frac{n(n+1)}{2}-n$$

7. Find the sum of n terms of the sequence {a_n} defined by a_n = 1/n.

$$S_n=\sum _{i=1}^n\frac{1}{i}\approx \ln n+\gamma$$

8. Find the sum of n terms of the sequence {a_n} defined by a_n = n^2 - 1.

$$S_n=\sum _{i=1}^n(i^2-1)=\frac{n(n+1)(2n+2)}{6}$$