Question 1: Sum the following series up to \(n\) terms.
Har part mein pehle \(T_n\) (nth term) nikalo, phir \(\sum n, \sum n^2, \sum n^3\) ke standard formulas lagao.
(i) \(1 \times 3 + 2 \times 5 + 3 \times 7 + \dots\)
\[
T_n = n(2n+1) = 2n^2 + n \implies S_n = \frac{n(n+1)(4n+5)}{6}
\]
(ii) \(1 \times 5 + 2 \times 8 + 3 \times 11 + \dots\)
\[
T_n = n(3n+2) = 3n^2 + 2n \implies S_n = \frac{n(n+1)(2n+3)}{2}
\]
(iii) \(1 \times 2 + 2 \times 5 + 3 \times 8 + \dots\)
\[
T_n = n(3n-1) = 3n^2 - n \implies S_n = n^2(n+1)
\]
(iv) \(1 \times 3 + 2 \times 4 + 3 \times 5 + \dots\)
\[
T_n = n(n+2) = n^2 + 2n \implies S_n = \frac{n(n+1)(2n+7)}{6}
\]
(v) \(1 \times 2 \times 4 + 2 \times 3 \times 7 + 3 \times 4 \times 10 + \dots\)
\[
T_n = n(n+1)(3n+1) = 3n^3 + 4n^2 + n \implies S_n = \frac{n(n+1)(9n^2+25n+14)}{12}
\]
(vi) \(2^2 + 4^2 + 6^2 + \dots\)
\[
T_n = 4n^2 \implies S_n = \frac{2n(n+1)(2n+1)}{3}
\]
(vii) \(3^2 + 6^2 + 9^2 + \dots\)
\[
T_n = 9n^2 \implies S_n = \frac{3n(n+1)(2n+1)}{2}
\]
(viii) \(4 \times 1^2 + 7 \times 2^2 + 10 \times 3^2 + \dots\)
\[
T_n = (3n+1)n^2 = 3n^3 + n^2 \implies S_n = \frac{n(n+1)(9n^2+13n+2)}{12}
\]
(ix) \(3 + (3+7) + (3+7+11) + \dots\)
\[
T_n = \sum_{k=1}^{n}(4k-1) = 2n^2 + n \implies S_n = \frac{n(n+1)(4n+5)}{6}
\]
(x) \(1^2 + (1^2+2^2) + (1^2+2^2+3^2) + \dots\)
\[
T_n = \frac{n(n+1)(2n+1)}{6} \implies S_n = \frac{n(n+1)^2(n+2)}{12}
\]
Answer: (i) \(\frac{n(n+1)(4n+5)}{6}\), (ii) \(\frac{n(n+1)(2n+3)}{2}\), (iii) \(n^2(n+1)\), (iv) \(\frac{n(n+1)(2n+7)}{6}\), (v) \(\frac{n(n+1)(9n^2+25n+14)}{12}\), (vi) \(\frac{2n(n+1)(2n+1)}{3}\), (vii) \(\frac{3n(n+1)(2n+1)}{2}\), (viii) \(\frac{n(n+1)(9n^2+13n+2)}{12}\), (ix) \(\frac{n(n+1)(4n+5)}{6}\), (x) \(\frac{n(n+1)^2(n+2)}{12}\)