3. a) Write the sequence got by adding one to the square of consecutive natural numbers starting from 1
b) What is the 10th term of this sequence?
c) Write the algebraic form of this sequence?
Answers
Answer:
n(n+1)(2n+1)b)The first term is
−
17
and the common difference is
3
.
Step-by-step explanation:
The formula to find sum of first n terms of a sequence of squares of natural numbers =
6
n(n+1)(2n+1)
The sum of the first
n
terms of an arithmetic sequence
a
1
,
a
2
,
a
3
,
...
with first term
a
1
and common difference
d
(i.e.
a
k
=
a
1
+
d
(
k
−
1
)
)
is given by
n
∑
k
=
1
a
k
=
n
(
a
1
+
a
n
2
)
As the sum of the first ten terms is
−
35
that gives us
10
(
a
1
+
a
10
2
)
=
−
35
As the tenth term is
10
that gives us
10
(
a
1
+
10
2
)
=
−
35
⇒
a
1
+
10
=
−
7
⇒
a
1
=
−
17
Which gives us the first term as
−
17
. To find the difference, we note that
10
=
a
10
=
a
1
+
9
d
=
−
17
+
9
d
⇒
9
d
=
27
⇒
d
=
Answer:
Given : sequence got by adding 1 to the square of consecutive natural numbers starting from 1
To Find : the sequence
10th term of this sequence
the algebraic form of this sequence
Solution:
adding 1 to the square of consecutive natural numbers starting from 1
Tₙ = n² + 1
n = 1
=> T₁= 1² + 1 = 2
n = 2
=> T₂= 2² + 1 = 5
n = 3
=> T₃= 3² + 1 = 10
n = 4
=> T₄= 4² + 1 = 17
n = 5
=> T₅= 5² + 1 = 26
n = 6
=> T₆= 6² + 1 = 37
Hence sequence is
2 , 5 , 10 , 17 , 26 , 37 and so on
Tₙ = n² + 1
n = 10
=> T₁₀ = 10² + 1 = 101
the algebraic form of this sequence n² + 1 n∈ natural numbers
Step-by-step explanation ;