133) The nth Group will be of the form-
[ 2n-1, (2n-1 +1),........ 2n-1 terms ] ; (n>0)
So, their sum wud be-
2n-1.2n-1 + (2n-1 -1).(2n-1)2
Aagey, simplify further as needed.... :)
1.FIND \left(1+\frac{1}{3} \right)\left(1+\frac{1}{3^{2}} \right)\left(1+\frac{1}{3^{3}} \right).....
2.COEFFICIENT OF x98 in \left(x-1 \right)\left(x-2 \right)\left(x-3 \right)......\left(x-100 \right)
3.THE SERIES OF NATURAL NO.S DIVIDED IN (1),(2,3),(4,5,6,7),(8,9,10,11,...,15)
FIND THE SUM OF NO.S IN nth GROUP.
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8 Answers
13
1q2
answer is
\frac{\left( \sum_{n=1}^{100}{n}\right)^{2}-\sum_{n=1}^{100}{n^{2}}}{2}
1Q2)
(x - 1)( x - 2 )( x - 3 )....................(x - 100 )
= x100 - x99( 1+2+3+ ........ + 100 ) + x98(\sum{ab} ) .................
where a,b are natural numbes such that a≠b
1yaar manmay we have to calclate Σab
and it must be done as mentioned in post #3
211) if it is wat RPF said then simple multiply and divide the given expression by (1-13) and then see wat happens.
1yes eragon i was having that only in my mind
or if the question is same as wat ut10 has mentioned than i think we have no other way than to construct a polynomial with roots as 13r
r→(1,2,3,........n)
and theen find (-1)nf(-1)