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68. The given series is $\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+?$

So, an = $\frac{1}{\left(n^{}1,2,3\dots \right)}\times \frac{1}{\left(n^{ }ten 2,3,4\dots \right)}$

$=\frac{1}{[1+(n-1)1]}\times \frac{1}{[2+(n-1)1]}$

$=\frac{1}{(1+x-1)}\times \frac{1}{(2+n-1)}$

$=\frac{1}{n(n+1)}$

$=\frac{(n+1)-n}{n(n+1)}$

$=\frac{(n+1)}{n(n+1)}-\frac{n}{n(n+1)}$

$a_n=\frac{1}{n}-\frac{1}{n+1}$

So. Putting n = 1, 2, 3….n.

a₁ = $\frac{1}{1}-\frac{1}{2}$

a₂ = $\frac{1}{2}-\frac{1}{3}$

a₃ = $\frac{1}{3}-\frac{1}{4}\dots .$

So, adding. L.H.S and R.H.S. up ton terms

a₁ + a² + a₃ + … + a_n = $\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+?\frac{1}{n}\right]-[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+?·\frac{1}{n}+\frac{1}{n+1}]$

 S_n = 1 $\frac{1}{n+1}$ { equal terns cancelled out}

 S_n = $\frac{n+1-1}{.n-1}$

 S_n = $\frac{n}{n+1}$

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67. The given series is 3 × 1² + 5 × 2² + 7 × 3² + ….

So,a_n = (n^th term of A P 3, 5, 7, .) (n^th term of A P 1, 2, 3, ….)²

a = 3, d = 5 -3 = 2a = 1, d = 2 -1 = 1.

= [3 + (n- 1) 2] [1 + (n- 1) 1]²

=[3 + 2n- 2] [1 + n- 1]²

(2n + 1)(n)²

= 2n³ + n²

So, = 5n2∑n³ + ∑n²

$=2·{\left[\frac{n^{}(n+1)}{2}\right]}^2+\left[\frac{n(n+1)(2n+1)}{6}\right]$

$=\frac{n(n+1)}{2}\left[2·\frac{n(n+1)}{2}+\frac{(2n+1)}{3}\right]$

$=\frac{n(n+1)}{2}\left[\frac{3n(n+1)+(2n+1)}{3}\right]$

$=\frac{n(n+1)}{2}\left[\frac{3n^2+3n+2n+1}{3}\right]$

$=\frac{n}{2}(n+1)·\left[\frac{3n^2+5n+1}{3}\right]$

$\frac{n(n+1)\left(3n^2+5n+1\right)}{6}$

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SGT University select students for admission in BTech courses on merit-basis as per the official website. Aspirants performance in Class 12 or equivalent exam is evaluated. Moreover, the university may also consider CUET scores for selection. The applicants further must meet the eligibility requirements for admission in the UG programme.

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66. Given series is 1× 2× 3 + 2× 3 ×4 + 3× 4 ×5 + … to n term

a_n = (n^th term of A. P. 1, 2, 3, …) ´× (n^th terms of A. P. 2, 3, 4) ×

i e, a = 1, d = 2- 1 = 1i e, a = 2, d = 3- 2 = 1

(nth term of A. P. 3, 4, 5)

i e, a = 3, d = 3 -4 = 1.

= [1 + (n -1) 1] ×[2 + (n -1):1]× [3 + (n- 1) 1]

= (1 + n -1)×(2 + n -1)×(3 + n -1)

= n (n + 1)(n + 2)

= n(n² + 2n + n + 2)

=n³ + 2n² + 2n.

S_n = ∑n³ + 3 ∑n² + 2 ∑n

$={\left[\frac{m(n+1)}{2}\right]}^2+\frac{3n(n+1)(2n+1)}{6}+\frac{2n(n+1)}{2}$

= $\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{3}{3}(2n+1)+2\right]$

= $\frac{n(n+1)}{2}\left[\frac{n^2+n}{2}+2n+1+2\right]$

$=\frac{n(n+1)}{2}\left[\frac{n^2+n+2\times 2n+2\times 3}{2}\right]$

$=\frac{n(n+1)}{2}\left[\frac{x^2+n+4x+6}{2}\right]$

$=\frac{n(n+1)}{2}\times \frac{\left[n^2+5n+6\right]}{2}$

$=\frac{n(n+1)}{4}\left[n^2+2n+3n+6\right]$

$=\frac{n(n+1)}{4}[n(n+2)+3(n+2)]$

$=\frac{n(n+1)(n+2)(n+3)}{4}$

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Disclaimer: This information is sourced from the official website and may vary.

65. Given series is 1×2+2 ×3+3× 4+4× 5+…

So, an (nth term of A.P 1, 2, 3…) (nth term of A.P. 2, 3, 4, 5…)

i e, a = 2, d = 2 -1 = 1i e, a = 2, d = 3 - 2 = 1

= [1 + (n- 1) 1] [2 + (n -1) 1]

= [1 + n- 1] [2 + n -1]

= n (n -1)

= n²-n.

S_n (sum of n terms of the series) = ∑n² + ∑n.

S_n = $\frac{n(n+1)(2n+1)}{6}$ + $\frac{n(x+1)}{2}$

= $\frac{n(n+1)}{2}\left[\frac{2n+1}{3}+1\right]$

= $\frac{n(n+1)}{2}\left[\frac{2n+1+3}{3}\right]$

$=\frac{n(n+1)}{2}\times \frac{(2n+4)}{3}$

$=\frac{n(n+1)\times 2(n+2)}{2\times 3}$

$=\frac{n(n+1)(n+2)}{3}$

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