**Find the sum of n terms of the series 1×2 + 2×3 + 3×4 + 4×5 +…**

Solution:

The given series is 1×2 + 2×3 + 3×4 + 4×5 +… n^{th} term, a_{n}=n(n+1)

**Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5+ …**

Solution:

The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + … n th term,

a_{n} = n ( n + 1) ( n + 2)

= (n^{2} + n) (n + 2)

= n^{3} + 3n^{2} + 2n

**Find the sum to n terms of the series 3 × 12 + 5 × 22 + 7 × 32 + …**

Solution:

The given series is 3 ×12 + 5 × 22 + 7 × 32 + … nth term,

a_{n} = ( 2n + 1) n^{2} = 2n^{3} + n^{2}

**Find the sum to n terms of the series**^{1}/_{1×2}+^{1}/_{2×3}+^{1}/_{3×4}+ …

Solution:

The given series is ^{1}/_{1×2} + ^{1}/_{2×3} + ^{1}/_{3×4} + …

n^{th} term, a_{n} = ^{1}/_{n(n+1)} = ^{1}/_{n} – ^{1}/_{n+1}

a_{1} = ^{1}/_{1} – ^{1}/_{2}

a_{2} = ^{1}/_{2} – ^{1}/_{3}

a_{3} = ^{1}/_{3} – ^{1}/_{4}…

Adding the above terms column wise, we get

a_{1} + a_{2} + a_{3} + … +a_{n} = [^{1}/_{1} + ^{1}/_{2} + ^{1}/_{3} +…+^{1}/_{n}] – [^{1}/_{2} + ^{1}/_{3} +…+^{1}/_{n+1}]

S_{n} = 1 – ^{1}/_{n+1 }= ^{n+1-1}/_{n+1} = ^{n}/_{n+1}

**Find the sum to n terms of the series 5**^{2}+ 6^{2}+ 7^{2 }+…+20^{2}

Solution:

The given series 5^{2} + 6^{2} + 7^{2 }+…+20^{2} nth terms

a_{n} = (n+4)^{2} = n^{2} + 8n + 16

= 1496 +1088+256

=2840

Therefore, 5^{2} + 6^{2} + 7^{2} + …+20^{2} = 2840

**Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…**

Solution:

The given series is 3 × 8 + 6 × 11 + 9 × 14 + … a_{n}

= (n^{th} term of 3, 6, 9 …) × (n^{th} term of 8, 11, 14 …)

= (3n) (3n + 5)

= 9n^{2} + 15n

**Find the sum to n terms of the series 1**^{2}+ (1^{2}+ 2^{2})+ (1^{2}+ 2^{2}+3^{2})+…

Solution:

The given series 1^{2} + (1^{2} + 2^{2})+ (1^{2} + 2^{2}+3^{2})+… a_{n} = 1^{2} + 2^{2}+3^{2}+…+n^{2}

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**Find the sum to n terms of the series whose n**^{th}term is given by n (n + 1)(n + 4).

Solution:

a_{n} = n (n + 1) (n + 4) = n(n^{2 }+ 5n + 4) = n^{3} + 5n^{2} + 4n

- Find the sum to n terms of the series whose n
^{th}terms is given by n^{2}+ 2^{n}

Solution:

a_{n} = n^{2} + 2^{n}

**Find the sum to n terms of the series whose n**^{th}terms is given by (2n – 1)^{2}

Solution:

a_{n} = (2n – 1)^{2} = 4n^{2} – 4n + 1