Question
Question: Find out the sum of the series \( \dfrac{1}{{1 \times 2}}{}^{25}{C_0} + \dfrac{1}{{2 \times 3}...
Find out the sum of the series
1×2125C0+2×3125C1+3×4125C2+............+26×27125C25 a. 26×27227−1 b. 26×27227−28 c. 21(26×27226−1) d. 52226−1
Solution
Hint: - Use(1+x)25=25C0+25C1x+25C2x2+.................+25C25x25, then apply integration on both sides with limit 0 to x
According to Binomial Theorem the expansion of (1+x)25is
(1+x)25=25C0+25C1x+25C2x2+.................+25C25x25...........(1)
Integrate equation 1 w.r.t.xWith limit 0 to x
0∫x(1+x)25dx=0∫x(25C0+25C1x+25C2x2+.................+25C25x25)dx
Let, (1+x)=t...................(2)
So when
x=0⇒t=1+0=1 x=x⇒t=1+x
Now, differentiate equation 2 w.r.t.x
⇒0+dx=dt⇒dx=dt
Substitute these value in equation 1
1∫1+xt25dt=0∫x(25C0+25C1x+25C2x2+.................+25C25x25)dx
Now integrate this equation as you know ∫tndt=[n+1tn+1]
⇒[26t26]11+x=[25C0x+25C12x2+25C23x3+.................+25C2526x26]ox
Now apply integrating limit
Now again integrate equation 3 w.r.t.xFrom limit 0 to 1.
0∫1[26(1+x)26−261]dx=0∫1[25C0x+25C12x2+25C23x3+.................+25C2526x26]dx
Let, (1+x)=t...................(4)
So when
x=0⇒t=1+0=1 x=1⇒t=1+1=2
Now, differentiate equation 4 w.r.t.x
⇒0+dx=dt⇒dx=dt
Substitute these value in equation 3
1∫2[26t26−261]dt=0∫1[25C0x+25C12x2+25C23x3+.................+25C2526x26]dx
Now integrate this equation as you know ∫tndt=[n+1tn+1]
[27×26t27−261]12=[25C02x2+25C12×3x3+25C23×4x4+.................+25C2526×27x27]01
Now apply integrating limit
[27×26227−261−(27×26127)]=[25C021+25C12×31+25C23×41+.................+25C2526×271−0−0−0] ⇒[27×26227−28]=1×2125C0+2×3125C1+3×4125C2+............+26×27125C25
Hence, option (b) is correct.
Note: - Whenever we face such type of problem the key concept we have to remember is that always remember the Binomial expansion of(1+x)n, then integrate the expansion w.r.t.x With limit 0 to x, then again integrate w.r.t.xwith limit 0 to 1, we will get the required answer.