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riya srivastava Grade: 11
        
Find the solution of this given series
1+5+11+19+29+....to n terms
one year ago

Answers : (2)

jagdish singh singh
168 Points
										
\hspace{-0.70cm}$ Let $S_{n} = 1+5+11+19+29+\cdots n$ terms\\\\\\ So $S_{n} = (1^2+0)+(2^2+1)+(3^2+2)+(4^2+3)+\cdots+(n^2+n-1)$\\\\\\ So $S_{n} = \sum^{n}_{r=1}(r^2+r-1) = \sum^{n}_{r=1}r^2+\sum^{n}_{r=1}r-\sum^{n}_{r=1}1$\\\\\\ So $S_{n} = \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}-n.$
one year ago
moumi roy
78 Points
										
 
S=1+5+11+19....+TN -S= -1-5-11.....-T(N-1)-N ADDING 0=1+4+6+8+....-T(N-1)-TN TN=1+4+6+8+10...T(N-1) TN=1+\frac{(N-1)(8+2N-4)}{2}=1+(N-1)(N+2)=N^{2}+N-1 \sum (TN)=\sum N^{2}+\sum N-N=\frac{N(N+1)(2N+1)}{6}+\frac{N(N+1)}{2}-N
DONE THROUG DIFFERENCE METHOD THERE IS A LITTLE PROBLEM IN THE DISPLAY.SRRY.HOPE YOU UNDERSTAND.NOT ACCEPTING LESS THAN 100
one year ago
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