- 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 +… nth term, an=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,
an = n ( n + 1) ( n + 2)
= (n2 + n) (n + 2)
= n3 + 3n2 + 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,
an = ( 2n + 1) n2 = 2n3 + n2
- 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 + …
nth term, an = 1/n(n+1) = 1/n – 1/n+1
a1 = 1/1 – 1/2
a2 = 1/2 – 1/3
a3 = 1/3 – 1/4…
Adding the above terms column wise, we get
a1 + a2 + a3 + … +an = [1/1 + 1/2 + 1/3 +…+1/n] – [1/2 + 1/3 +…+1/n+1]
Sn = 1 – 1/n+1 = n+1-1/n+1 = n/n+1
- Find the sum to n terms of the series 52 + 62 + 72 +…+202
Solution:
The given series 52 + 62 + 72 +…+202 nth terms
an = (n+4)2 = n2 + 8n + 16
= 1496 +1088+256
=2840
Therefore, 52 + 62 + 72 + …+202 = 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 + … an
= (nth term of 3, 6, 9 …) × (nth term of 8, 11, 14 …)
= (3n) (3n + 5)
= 9n2 + 15n
- Find the sum to n terms of the series 12 + (12 + 22)+ (12 + 22+32)+…
Solution:
The given series 12 + (12 + 22)+ (12 + 22+32)+… an = 12 + 22+32+…+n2
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- Find the sum to n terms of the series whose nth term is given by n (n + 1)(n + 4).
Solution:
an = n (n + 1) (n + 4) = n(n2 + 5n + 4) = n3 + 5n2 + 4n
- Find the sum to n terms of the series whose nth terms is given by n2 + 2n
Solution:
an = n2 + 2n
- Find the sum to n terms of the series whose nth terms is given by (2n – 1)2
Solution:
an = (2n – 1)2 = 4n2 – 4n + 1
