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  • If  <img alt="\mathrm{S}_{\mathrm{n}}=4+11+21+34+50+\ldots" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BS%7D_%7B%5Cmathrm%7Bn%7D%7D%3D4&plus;11&plus;21&plus;34

If  \mathrm{S}_{\mathrm{n}}=4+11+21+34+50+\ldots . to  \mathrm{n} terms, then \frac{1}{60}\left(\mathrm{~S}_{29}-\mathrm{S}_9\right)  is equal to

Option: 1

220


Option: 2

227


Option: 3

226


Option: 4

223


Answers (1)

best_answer

S_n=4+11+21+34+50+\ldots+n terms

Difference are in A.P .

Let T_n=a n^2+b n+c

\begin{aligned} & T_1=a+b+c=4 \\ & T_2=4 a+2 b+c=11 \\ & T_3=9 a+3 b+c=21 \end{aligned}

By solving these 3 equations

a=\frac{3}{2}, b=\frac{5}{2}, c=0

So T_n=\frac{3}{2} n^2+\frac{5}{2} n

\begin{aligned} & S_n=\Sigma T_n \\ & =\frac{3}{2} \sum n^2+\frac{5}{2} \sum n \\ & =\frac{3}{2} \frac{n(n+1)(2 n+1)}{6}+\frac{5}{2} \frac{(n)(n+1)}{2} \\ & =\frac{n(n+1)}{4}[2 n+1+5] \\ & S_n=\frac{n(n+1)}{4}(2 n+6)=\frac{n(n+1)(n+3)}{2} \\ & \frac{1}{60}\left(\frac{29 \times 30 \times 32}{2}-\frac{9 \times 10 \times 12}{2}\right)=223 \end{aligned}

Posted by

vishal kumar

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