club. Write a molice to senion be put up the to assemble studant! S ab. You are the secretary of the school sponts 2. motice board intimating selection of the school team for the • for practise for the selection. for the district games. Mention the date and time.
please it's urgent,hurry up?
Answers
Answer:
Consider,
\begin{gathered}\rm \: 1 + (1 + 2) + (1 + 2 + 3) + - - - + (1 + 2 + 3 + - - - + n) \\ \end{gathered}
1+(1+2)+(1+2+3)+−−−+(1+2+3+−−−+n)
Let assume that,
\begin{gathered}\rm \:S_n = 1 + (1 + 2) + (1 + 2 + 3) + - - - + (1 + 2 + 3 + - - - + n) \\ \end{gathered}
S
n
=1+(1+2)+(1+2+3)+−−−+(1+2+3+−−−+n)
So, nth term of the series is
\begin{gathered}\rm \: T_n = 1 + 2 + 3 + - - + n \\ \end{gathered}
T
n
=1+2+3+−−+n
\begin{gathered}\rm \: T_n = \displaystyle\sum_{k=1}^n\rm k \\ \end{gathered}
T
n
=
k=1
∑
n
k
\begin{gathered}\rm \: T_n = \dfrac{n(n + 1)}{2} \\ \end{gathered}
T
n
=
2
n(n+1)
\begin{gathered}\rm \: T_n = \dfrac{1}{2}( {n}^{2} + n) \\ \end{gathered}
T
n
=
2
1
(n
2
+n)
So, Sum of n terms of series is given by
\begin{gathered}\rm \: S_n = \displaystyle\sum_{k=1}^n\rm T_k \\ \end{gathered}
S
n
=
k=1
∑
n
T
k
\begin{gathered}\rm \: S_n = \dfrac{1}{2} \displaystyle\sum_{k=1}^n\rm ( {k}^{2} + k) \\ \end{gathered}
S
n
=
2
1
k=1
∑
n
(k
2
+k)
\begin{gathered}\rm \: S_n = \dfrac{1}{2}\bigg( \displaystyle\sum_{k=1}^n\rm {k}^{2} + \displaystyle\sum_{k=1}^n\rm k\bigg) \\ \end{gathered}
S
n
=
2
1
(
k=1
∑
n
k
2
+
k=1
∑
n
k)
\begin{gathered}\rm \: S_n = \dfrac{1}{2}\bigg(\dfrac{n(n + 1)}{2} + \dfrac{n(n + 1)(2n + 1)}{6} \bigg) \\ \end{gathered}
S
n
=
2
1
(
2
n(n+1)
+
6
n(n+1)(2n+1)
)
\begin{gathered}\rm \: S_n = \dfrac{1}{2} \times \dfrac{n(n + 1)}{2}\bigg(1 + \dfrac{2n + 1}{3} \bigg) \\ \end{gathered}
S
n
=
2
1
×
2
n(n+1)
(1+
3
2n+1
)
\begin{gathered}\rm \: S_n = \dfrac{1}{2} \times \dfrac{n(n + 1)}{2}\bigg(\dfrac{3 + 2n + 1}{3} \bigg) \\ \end{gathered}
S
n
=
2
1
×
2
n(n+1)
(
3
3+2n+1
)
\begin{gathered}\rm \: S_n = \dfrac{1}{2} \times \dfrac{n(n + 1)}{2}\bigg(\dfrac{2n + 4}{3} \bigg) \\ \end{gathered}
S
n
=
2
1
×
2
n(n+1)
(
3
2n+4
)
\begin{gathered}\rm \: S_n = \dfrac{1}{2} \times \dfrac{n(n + 1)}{2}\bigg(\dfrac{2(n + 2)}{3} \bigg) \\ \end{gathered}
S
n
=
2
1
×
2
n(n+1)
(
3
2(n+2)
)
\begin{gathered}\rm\implies \:S_n = \dfrac{n(n + 1)(n + 2)}{6} \\ \end{gathered}
⟹S
n
=
6
n(n+1)(n+2)
Hence, Proved